◄Topic 5Unit 4 Work, Energy and Power►

Unit 3 : Moments and Equilibrium of Bodies

  • Unit 3 : Moments and Equilibrium of Bodies

    Key Unit Competence 

    Analyse the principle of moments and equilibrium of bodies. .

    My goals

    By the end of this unit, I should be able to:

    * Explain the principle of moments and apply it to equilibrium of a 
    body.
    * Come out with the effects of forces when applied onto a body.

    * Know the effects of forces.

    INTRODUCTORY ACTIVITY

    When a patient goes to hospital having a headache and fever, a doctor may 
    require a blood test for malaria. When a sample of blood is taken, it is not 
    possible to check whether a patient has malaria or not. But a laboratory 
    technician may need to test the blood using some instrument and decide 

    whether the patient has malaria or not.

    Questions:
    (i) Which instrument do you think may be used to test malaria from 
    blood sample?
    (ii) In summary, discuss how that instrument function.

    (iii) What other instrument do you think can be used for such purpose?

    Introduction

    In here, we shall majorly concetrate on the turning effect of force. As you 
    know, it is very hard to close a door when you apply force near its turning 
    point. That’s why door handles are always put at the end of the door so that 
    the distance from the turning point to where force is applied increases. This 
    increases the turning effect of the force applied. Which is the effect of forces 

    on bodies one of our interest in this unit.

    Scalar and vector quantities

    Activity 1

    Try to stand bricks in a line behind one another. Push one brick.
    (i) What happens to other bricks?

    (ii) What if in the process one brick stops, what would happen?

    In daily life, we normally pull the objects from one place to another. When 
    pulling a goat that is to be tethered, obviously it will take the direction of the 
    pull. We can call this a force. This is a quantity that changes body’s state of 

    rest or uniform motion.

    You noticed that after pushing your friend he/she changed position and 
    direction. Hence, a force has both magnitude and direction. This quantity can 
    be termed as a vector quantity. This is a quantity with both magnitude and 

    direction.

    Activity 2

    (i) Using the above example, discuss in groups or as a class other 
    vector Quantities.
    (ii) Analyse the effects of these physical quantities.
    (iii) In daily life, how are these quantities utilised? 

    (iv) Ask your friend what time is it? 

    You will realise that he/she will tell the exact time not even indicating direction. 

    Such a quantity is termed to be a scalar quantity.

    A scalar quantity is a physical quantity that is defined by only magnitude 

    (size).

    Other examples of scalar quantities are volume, mass, speed, and time 
    intervals. The rules of ordinary arithmetic are used to manipulate scalar 

    quantities.

    Application activity 3.1

    Of the following physical quantities, group them in different 
    sets of scalar and vector quantities: mass, energy, power, weight, 
    acceleration, velocity, momentum, time, impulse, magnetic flux density, 

    pressure, displacement.

    Force as vector

    Activity 3

    1. As an individual or a group push the desk.
    2. What happens to it?
    3. What causes the change in position?
    a) Let as a class move to:
    (i) Football pitch.
    (ii)Net ball pitch.
    (iii) Basket ball play ground.
    Try to kick a ball. What happens to it? What causes it to change its 
    position?

    Note what you observe.

    Also, as you sit reading this book, you eventually feel tired. This is because of 

    gravitational force acting on your body and yet you remain stationary.

    From the above examples, we can define the “quantity force”. We have to 
    know the direction and the magnitude. For that matter, we conclude that the 
    force is vector quantity. We can think of force as that which causes an object 

    to accelerate.

    What happens when several forces act simultaneously on an object? In this case, 
    the object accelerates only if the net force acting on it is not equal to zero. The 
    net force acting on an object is defined as the vector sum of all forces acting on 
    the object. (We sometimes refer to the net force as the total force, the resultant 

    force, or the unbalanced force.)

    Application activity 3.2

    1. ............................ is an example of a scalar quantity
    a) Velocity.
    b) Force.
    c) Volume.

    d) Acceleration.

    2. ............................ is an example of a vector quantity
    a) Mass.
    b) Force.
    c) Volume.

    d) Density.

    3. A scalar quantity:
    a) always has mass.
    b) is a quantity that is completely specified by its magnitude.
    c) shows direction.

    d) does not have units.

    4. A vector quantity
    a) can be a dimensionless quantity.
    b) specifies only magnitude.
    c) specifies only direction.

    d) specifies both a magnitude and a direction.

    5. A boy pushes against the wall with 50 kilogrammes of force. The wall 
    does not move. The resultant force is:
    a) -50 kilogrammes.
    b) 100 kilogrammes.
    c) 0 kilogrammes.

    d) -75 kilogrammes.

    6. A man walks 3 miles north then turns right and walks 4 miles east. The 
    resultant displacement is:
    a) 1 kilometre SW
    b) 7 kilometres NE
    c) 5 kilometres NE

    d) 5 kilometres E

    7. A plane flying 500km/hr due north has a tail wind of 45 mi/hr the 
    resultant velocity is:
    a) 545 kilometres/hour due south.
    b) 455 kilometres/hour north.
    c) 545 kilometres/hour due north.

    d) 455 kilometres/hour due south.

    8. The difference between speed and velocity is:
    a) Speed has no units.
    b) Speed shows only magnitude, while velocity represents both 
    magnitude (strength) and direction.
    c) They use different units to represent their magnitude.

    d) Velocity has a higher magnitude.

    9. The resultant magnitude of two vectors
    a) Is always positive.
    b) Can never be zero.
    c) Can never be negative.

    d) Is usually zero.

    10.Which of the following is not true.
    a) Velocity can be negative.
    b) Velocity is a vector.
    c) Speed is a scalar.

    d) Speed can be negative.

    Table summarising Scalar and vector Quantities

    V

    Turning effect of force

    Moment of a force about a point
    Every time we open a door, turn on a tap or tighten up a nut with a spanner, 
    we exert a turning force. The combined effect of the force and distance which 
    determines the magnitude of the turning force is called the moment of the 

    force or torque and is defined as follows: 

    “The moment (turning effect) of a force about a point is the force multiplied 
    by the perpendicular distance from the place where the force is applied to that 

    point.” Fig. a

    C

    Or Moment is force times lever arm where is the lever arm, and the 
    perpendicular symbol egg reminds us that we must use the distance from the 
    axis of rotation that is perpendicular to the line of action of the force (Fig.3.1 

    a). The SI unit for moment is N m.

    A lever arm or moment arm is the perpendicular distance from the axis of 

    rotation to a line drawn along the direction of the force.

    C

    An equivalent way of determining the torque associated with a force is to 
    resolve the force into components parallel and perpendicular to the line that 
    connects the axis to the point of application of the force, as shown in Fig. 3.1b. 
    The component exerts no torque since it is directed at the rotation axis (its 
    lever arm is zero). Hence the torque will be equal to times the distance r from 

    the axis to the point of application of the force:

    C

    Activity 4

    (i) Suspend a meter rule at its middle point, either by passing a string 
    through a hole or a knife edge. If necessary stick plasticine one one 
    and to make it balance exactly.

    (ii) Tie loops of thread to several 

    (iii) Hang a 0.5 N weight A (the load) one the left hand side of the ruler 
    on the 34 cm mark (16 cm ) from the fulcrum)

    (iv) Place another weight 0.2 N (the effort) on the other side and move 

    it until the ruler balances.

    C

    (v) Note the distances of the weights from the fulcrum, i.e from the 
    midpoint.

    (vi) Record the results in a suitable as shown in the table.

    (vii) Reapeat several times with (a) the same weights in different 

    position, and (b) different weights.

    C

    The table suggests that when the turning effects of forces acting 
    on an object are balanced the sum of clock wise moments is the 
    same as the sum of anticlockwise moments. This is the principle of 

    moment

    Principle of moment of force

    When a body is in equilibrium (balanced), the sum of the anticlockwise 
    moments about any point is equal to the sum of the clockwise moments about 

    the same point.

    C

    Example 

    In kinesiology (the study of human motion), it is often useful to know the 
    location of the center of mass of a person. This can be determined with 
    the arrangement shown fig A. A plank of weight 40 N is placed on two 
    scales separated by 2.0 m. A person lies on the plank and the left scale 
    reads 3214 N and the right scale reads 216 N as shown in fig B. What is 

    the distance from the left scale to the person’s center of mass?

    C

    C

    C

    A torque is a quantity that measures the ability of a force to rotate an 
    object around some axis. Net torque produces rotation. A torque is positive 
    or negative; depending on the direction the force tends to rotate an object. 

    Torques that produce counterclockwise rotation are defined to be positive.

    Example 

    A basketball is being pushed by two basketball players during tip-off. 
    Assuming each force acts perpendicular to the axis of rotation through 

    the centre of ball the ball; find the net torque acting on the ball.

    C

    C

    Torque should not be confused with work
    Torque (τ = Fd sin θ ) and work (W = Fd cos θ ) can both be expressed in 
    units of N m, so be careful to distinguish torque and work. The components of 
    a force that produces work is parallel to a distance (the displacement), while 
    the component of force that produces torque is perpendicular to a distance (the 

    lever arm). 

    Torque should not be confused with force. Forces can cause a change in 
    linear motion, as described by Newton’s second law. Forces can also cause a 
    change in rotational motion, but the effectiveness of the forces in causing this 
    change depends on both the forces and the moment arms of the forces, in the 

    combination that we call torque. 

    If two or more forces are acting on a rigid object, as shown in Fig.3.3, each 
    tends to produce rotation about the pivot at O. We use the convention that the 
    sign of the torque resulting from a force is positive if the turning tendency 
    of the force is counterclockwise and is negative if the turning tendency is 

    clockwise.

    C

    Equilibrium of a body

    Conditions for equilibrium

    Objects in daily life have at least one force acting on them (gravity). If they are 
    at rest, then there must be other forces acting on them as well so that the net 
    force is zero. A book at rest on a table, for example, has two forces acting on 
    it, the downward force of gravity and the normal force the table exerts upward 

    on it (Fig. 3.4).

    C

    Because the book is at rest, Newton’s second law tells us that the net force 
    on it is zero. Thus the upward force exerted by the table on the book must be 
    equal in magnitude to the force of gravity acting downward on the book. Such 
    an object is said to be in equilibrium (Latin for “equal forces” or “balance”) 
    under the action of these two forces. This is often called the first condition 

    for equilibrium (Fig.3.5a).

    Force not only push or pull but have a turning-effect or moment about an 
    axis.     In cases of equilibrium the moments have also to be considered. If forces 
    act at different points on an extended body an additional requirement must 
    be satisfied to ensure that the body has no tendency to rotate: The sum of the 
    torques about any point must be zero (Fig.3.5.b). This is called the second 

    condition for equilibrium.

    C

    C

    A rigid body is in mechanical equilibrium when the sum of all forces on all 
    particles of the system is zero (i.e. when all the particles of the system are 
    at rest or that its center of mass moves with constant velocity relative to the 
    observer and the total force on each particle is permanently zero)., and also 
    the sum of all torques on all particles of the system is zero so that its state of 
    rotational motion remains constant. The bodies are rigid if they do not deform 

    under the action of applied forces.

    We will apply the first and second conditions for equilibrium to situations 
    in which a rigid body is at rest (no translation or rotation). Such a body is 
    said to be in static equilibrium (Fig. 3.6). But the same conditions apply to 
    a rigid body in uniform translational motion (without rotation), such as an 
    airplane in flight with constant speed, direction, and altitude. Such a body is in 

    equilibrium but is not static.

    C

    The above conditions of equilibrium are also used to determine the resultant 
    of non-parallel, non-concurrent systems of forces i.e. all of the lines of 
    action of the forces in this system do not meet at one point. The parallel force 
    system was a special case of this type. Since all of these forces are not entirely 
    parallel, the position of the resultant can be established using the graphical or 

    algebraic methods of resolving co-planar forces 

    There are a number of ways in which one could resolve the force system that 
    is shown. One graphical method would be to resolve a pair of forces using the 
    parallelogram or triangle method into a resultant. The resultant would then be 
    combined with one of the remaining forces and a new resultant determined, 
    and so on until all of the forces had been accounted for. This could prove to 
    be very complex if there are a great number of forces. The algebraic solution 
    to this system would potentially be simpler if the forces that are applied to the 

    system are easy to break into components.

    ■ Addition of Forces in a Plane (Stevinus law)
    If all the forces acting on a body act in a plane, they are called coplanar forces. 
    If they have a common point of application they are called concurrent forces

    Consider a body that is subjected to two forces F1 and F2, whose lines of 
    action intersect at point A (Fig. 3.7). It is postulated that the two forces can 
    be replaced by a statically equivalent force R. This postulate is an axiom; it is 
    known as the parallelogram law of forces. The force R is called the resultant 
    of F1 and F2 . It is the diagonal of the parallelogram for which F1 and F2 are 

    adjacent sides. 

    C

    Now consider a system of n forces that all lie in a plane and whose lines of 
    action intersect at point A (Fig. 3.8). Such a system is called a coplanar system 
    of concurrent forces. The resultant can be obtained through successive 
    application of the parallelogram law of forces. Mathematically, the summation 

    may be written in the form of the following vector equation:

    C

    C

    ■ Representation in Cartesian Coordinates

    It is usually convenient to resolve forces into two components that are 
    perpendicular to each other. The directions of the components may then be 
    given by the axes x and y of a Cartesian coordinate system (Fig. 3.8). The 
    quantities Fx and Fy are called the coordinates of the vector F or components 

    of F.

    C

    Examples

    1. A hiker begins a trip by first walking 25.0 km southeast from her car. 
    She stops and sets up her tent for the night. On the second day, she 
    walks 40.0 km in a direction 60.0° north of east, at which point she 
    discovers a forest ranger’s tower.
    (a) Determine the components of the hiker’s displacement for each 
    day.
    (b)Determine the components of the hiker’s resultant displacement 
    R for the trip. Find an expression for R in terms of unit vectors.

    (c) Determine the magnitude and direction of the total displacement.

    Solution

    (a) If we denote the displacement vectors on the first and second days by 
    A and B, respectively, and use the car as the origin of coordinates, we 

    obtain the vectors shown in Figure below.

    C

    C

    2. Under what circumstances would a nonzero vector lying in the xy 

    plane have components that are equal in magnitude?

    Solution

    Any vector that points along a line at 45° to the x and y axes has components 

    equal in magnitude.

    3. In what circumstance is the x component of a vector given by the 

    magnitude of the vector times the sine of its direction angle?

    Solution

    If the direction of a vector is specified by giving the angle of the vector 
    measured clockwise from the positive y-axis, then the x-component of the 
    vector is equal to the sine of the angle multiplied by the magnitude of the 

    vector.

    4. If A = B, what can you conclude about the components of A and B?

    Solution

    Any vector that points along a line at 45° to the x and y axes has components 

    equal in magnitude.

    ■ Lami’s theorem

    Lami’s theorem gives the conditions of equilibrium for three forces acting at 
    a point O. Lami’s theorem states that if three forces acting at a point are in 
    equilibrium, then each of the force is directly proportional to the sine of the 

    angle between the remaining two forces.

    C

    Branch of mechanics which deals with state of equilibrium is called statics. 
    Statics is the branch of mechanics concerned with the analysis of loads 
    (force, torque/moment) on physical systems in static equilibrium, that is, in 
    a state where the relative positions of subsystems do not vary over time, or 
    where components and structures are at a constant velocity. When in static 
    equilibrium, the system is either at rest, or its center of mass moves at constant 
    velocity. The study of moving bodies is known as dynamics, and in fact the 

    entire field of statics is a special case of dynamics.

    Stability and Balance

    An object in static equilibrium, if left undisturbed, will undergo no translational 
    or rotational acceleration since the sum of all the forces and the sum of all the 
    torques acting on it are zero. However, if the object is displaced slightly, three 

    outcomes are possible:

    • Equilibrium is said to be stable if small, externally induced 
    displacements from that state produce forces that tend to oppose the 
    displacement and return the body or particle to the equilibrium state. 
    Examples include a weight suspended by a spring or a brick lying on a 

    level surface. 

    • Equilibrium is unstable if the least departure produces forces that tend 
    to increase the displacement. An example is a ball bearing balanced on 

    the edge of a razor blade.

    • Static Equilibrium (neutral equilibrium) is equilibrium where 
    all forces are balanced, but it also applies to bodies in uniform or 
    accelerated motion. For example, a book resting on a table applies a 
    downward force equal to its weight on the table. According to the third 
    law, the table applies an equal and opposite force to the book. This 
    force occurs because the weight of the book causes the table to deform 

    slightly so that it pushes back on the book like a coiled spring.

    By summing up

    • A system is said to be in stable equilibrium if, when displaced from 
    equilibrium, it experiences a net force or torque in a direction opposite 

    the direction of the displacement.

    • A system is in unstable equilibrium if, when displaced from equilibrium, 
    it experiences a net force or torque in the same direction as the 

    displacement from equilibrium.

    • A system is in neutral equilibrium if its equilibrium is independent of 

    displacements from its original position.

    C

    Centre of gravity and center of mass

    Concept of centre of gravity and center of mass

    The center of gravity is the average location of the weight of an object. The 
    centre of gravity is defined as the point of application of the resultant force 
    due to the earth’s attraction on it. The center of gravity is a geometric property 

    of any object.

    The centre of gravity of a body also coincides with its centre of mass. The 
    center of mass of an object may be defined as the point at which an applied 

    force produces acceleration but no rotation

    C

    Centre of gravity and base of support of a body

    1. For balance to exist, the line of gravity must intersect the base of support.

    2. If the area of the base of support of an object is increased, this tends to 
    increase the stability of the object.

    3. The lower the center of gravity is above the base of support the more 
    stable the object tends to be. (This is true even though the size of the 
    base of support is unchanged.)

    4. Objects that are more massive tend to be more stable.

    5. For an object, the farther the line of gravity’s intersection is from the 
    edge of its base of support the more stable the object tends to be in that 
    direction.

    Determining the center of gravity

    Determining the center of gravity is very important for any flying object. In 
    general, determining the center of gravity (cg) is a complicated procedure 
    because the mass (and weight) may not be uniformly distributed throughout the 
    object. If the mass is uniformly distributed, the problem is greatly simplified. 
    If the object has a line (or plane) of symmetry, the center of gravity lies on the 

    line of symmetry. 

    For a solid block of uniform material, the center of gravity is simply at the 

    average location of the physical dimensions. 

    Example

    C

    Activity 5

    To determine the centre of gravity of different regular shapes

    Apparatus

    * Manila paper, scissors, a pencil and a ruler

    Procedure

    • Make a number of shapes from a manila paper, as below:

    C

    • Find the centre of gravity of those different figures.

    Conclusion
    The point of intersection of diagonals (a,d and e), bisectors (c) , 

    diameters(b) is the centre of gravity of those figures

    For a general shaped object, there is a simple mechanical way to determine 

    the center of gravity: 

    In Step 1, you hang the object from any point and you drop a weighted string 

    (plumb line) from the same point. Draw a line on the object along the string.

    C

    For Step 2, repeat the procedure from another point on the object you now 
    have two lines drawn on the object which intersect. The center of gravity is 
    the point where the lines intersect. This procedure works well for irregularly 

    shaped objects that are hard to balance.

    Activity 6

    To determine the centre of gravity of an irregularly shaped lamina
    Apparatus

    * A plumb line, a thread, a stand and a cardboard

    C

    Procedure

    • Make 3 holes, A, B and C on the edges the cardboard.
    • Suspend it by a rod through the hole A as shown in Figure2.4 
    • Tie a plumb line on the rod beside the cardboard.
    • After the cardboard and plumb line have stopped swinging, draw 
    a vertical line on the cardboard as set by plumb line.
    • Repeat the experiment using other holes B and C. The lines 
    intersect at a point noted G
    • Balance the cardboard with the tip of a pencil at the point of 

    intersection of the three lines. What do you observe?

    Observation

    The suspended object will always rest with its centre of gravity vertically 
    below the point of support. The object balances on the tip of the pencil if 

    placed at its centre of gravit.

    C

    C

    C

    Stability and center of gravity

    Consider a ball suspended freely from a string is in stable equilibrium, for if it 
    is displaced to one side, it will return to its original position (Fig. 3.14a) due 
    to the net force and torque exerted on it. On the other hand, a pencil standing 
    on its point is in unstable equilibrium. If its center of gravity is directly over 
    its tip (Fig. 3.14b), the net force and net torque on it will be zero. But if it 
    is displaced ever so slightly as shown—say, by a slight vibration or tiny air 
    current—there will be a torque on it, and this torque acts to make the pencil 
    continue to fall in the direction of the original displacement. 
    Finally, an example of an object in neutral equilibrium is a sphere resting on a 
    horizontal tabletop. If it is moved slightly to one side, it will remain in its new 

    position—no net torque acts on it.

    C

    In general, an object whose center of gravity (CG) is below its point of support, 
    such as a ball on a string, will be in stable equilibrium. Consider a standing 
    refrigerator (Fig. 3.15a). If it is tipped slightly, it will return to its original 
    position due to the torque on it as shown in Fig. 3.15b. But if it is tipped too 
    far, Fig. 3.15c, it will fall over. The critical point is reached when the CG shifts 
    from one side of the pivot point to the other.When the CG is on one side, the 
    torque pulls the object back onto its original base of support, Fig. 3.15b. If the 
    object is tipped further, the CG goes past the pivot point and the torque causes 

    the object to topple, Fig. 3.15c.

    C

    In general, 

    • an object whose center of gravity is above its base of support will be 
    stable if a vertical line projected downward from the CG falls within the 
    base of support. This is because the normal force upward on the object 
    (which balances out gravity) can be exerted only within the area of 
    contact, so if the force of gravity acts beyond this area, a net torque will 
    act to topple the object. 
    • the larger the base and the lower the CG, the more stable the object.
    • an object tends to fall when its center of gravity is away from the base 

    that supports it. 

    Applications of equlilibrium

    Tower Crane – Method of Joints

    The tower crane shown in the figure below consists of tower DCE fixed at the 
    ground and two jibs AC and CB. The jibs are supported by tie bars AD and 
    DB, and are assumed to be attached to the tower by pinned connections. The 

    counterweight WC weighs and the crane has a lifting capacity of W.

    C

    Beam balance

    It consist of pivoted horizontal lever of equal length arms called the beam, 
    with a weighing pan also called scale, scale pan or boson, suspended from 

    each arm.

    C

    The unknown mass is placed on one pan and standard masses are added on the 
    other pan until is as close to equilibrium as possible. In precision balances a 
    slider mass is moved along a graduated scale. The slider position gives a fine 

    collection to the mass value.

    Example

    A uniform meter stick supported at the 25 cm mark is in equilibrium 
    when a 1 kg rock is suspended at the 0 cm end (as shown in Fig.). Is the 
    mass of the meter stick greater than, equal to, or less than the mass of the 

    rock? Explain your reasoning.

    C

    Solution

    C

    Application Activity 3.3

    A friend of yours that has been sick comes from the hospital where he 
    had gone for medication telling you that he is 62 kg and that is what is 
    indicated on his medical papers.
    Imagine you have the following
    * A strong wood bar of a bout 5m (or any length as long as you know 
    its actual length) and has a mass of 10kg.
    * A wood that you can use as a pivot

    a)Since you know the exact mass of your friend and that of the 
    wood. Explain all the procedures you can follow to know your 
    mass. In your procedures include all the necessary equations and 
    diagrams you may need to use.

    b)Do you think the mass (your mass) you obtained is accurate? 
    Explain your reasoning.

    c)What do you think may be the source of errors in determining 
    your mass? Or any unknown mass?

    d)Basing on your knowledge and understanding you have obtained 
    from this unit, how can this unit be applied in your career you are 

    pursuing?

    END UNIT ASSESSMENT

    1. The uniform bar shown below weighs 40 N and is subjected to the 
    forces shown. Find the magnitude, location, and direction of the force 

    needed to keep the bar in equilibrium.

    C

    2. System given below is in equilibrium. If the potential energies of 
    objects A and B are equal, find the mass of object A in terms of G. 

    (Rod is homogeneous and weight of it is G.)

    C

    A 172 cm tall person lies on a light (massless) board which is sup-ported by 
    two scales, one under the top of her head and one beneath 
    the bottom of her feet see Fig. The two scales read respectively 35.1 
    kg and 31.6 kg. What distance is the center of gravity of this person 

    from the bottom of her feet?

    C

    3. A) A seesaw consisting of a uniform board of mass M =10kg and 
    length l=2m supports a father and daughter with masses mf and md, 50 

    and 20kg respectively as shown in the Figure below.

    C

    The support (called the fulcrum) is under the center of gravity of the 
    board, the father is a distance d from the center, and the daughter is a 

    distance l / 2 from the center.

    a. Determine the magnitude of the upward force n exerted by the 

    support on the board.

    b. Determine where the father should sit to balance the system.

    B) Three children are trying to balance on a seesaw, which consists of 
    a fulcrum rock, acting as a pivot at the center, and a very light board 

    3.6 m long (see fig.).

    c

    Two playmates are already on either end. Boy A has a mass of 45 kg, and 
    girl B a mass of 35 kg. Where should girl C, whose mass is 25 kg, place 

    herself so as to balance the seesaw?

    4. A uniform 1500kg beam, 20m long, supports a 15,000kg printing press 
    5 from the right support column, see the figure. Calculate the force on 

    each of the vertical support columns.

    c

    5. A horizontal rod AB is suspended at its ends by two strings. (See the 
    figure below). The rod is 0.6m long and its weight of 3N acts at G 

    where AG is 0.4m and BG is 0.2m. Find the tensions X and Y

    c

    C

    C

    C

    ◄Topic 5Unit 4 Work, Energy and Power►