We discuss problems, which highlight certain aspects of the study leading to pulleys. The questions are categorized in terms of the characterizing features of the subject matter :

Problem 1 : A person is pulling a block of 50 kg at a uniform speed with the help of a pulley and rope arrangement as shown in the figure. If the person weighs 60 kg, find the normal force that the person exerts on the surface underneath.

Figure 1: Applied force is opposite to the direction of force lifting the block.

Change of force direction

Solution : In order to find normal force applied by the person on the surface underneath, we need to carry out force analysis of the external forces on the person. The surface and the person apply normal force on each other, which is equal in magnitude but opposite in direction. The forces on the person are (i) weight, “mg”, acting downward (ii) normal force, “N”, acting upward and (iii) tension, “T”, acting upward.

The free body diagram of the person is shown on the right side of the figure shown below.

Figure 2: Free body diagrams of block and boy are shown.

Free body diagrams

∑ F y ⇒ T + N = m g ∑ F y ⇒ T + N = m g

⇒ N = 60 X 10 − T = 600 − T ⇒ N = 60 X 10 − T = 600 − T

It is clear that we would need the value of tension, “T”, to solve the equation for the normal force. We can find the value of “T” by considering the FBD of the block being raised. The FBD of the block is shown on the left side of the figure above. Note that we consider that person is raising the block without acceleration (at uniform velocity as given in the question).

∑ F y = T − 50 g = 0 ∑ F y = T − 50 g = 0

⇒ T = 50 g = 50 X 10 = 500 N ⇒ T = 50 g = 50 X 10 = 500 N

Putting this value in the equation for normal force, we have :

⇒ N = 600 − T = 600 − 500 = 100 N ⇒ N = 600 − T = 600 − 500 = 100 N

Problem 4 : In the arrangement shown in the figure, the block “A” moves with a velocity 4 m/s towards right. The string and the pulleys are “mass-less” and friction is absent everywhere. What is relative velocity of block “B” with respect to “A”?

Figure 7: Pulley and block system.

Horizontal pulley system

Solution : The blocks are connected by a single string, which is intervened by two pulleys. We can connect velocities of blocks, if we have constraint relation for the length of string. First differentiation of the constraint relation will yield the required relation between velocities.

Since the pulleys are connected to blocks as their part (their relative positions do not change due to motion), we treat block and the attached pulley as a single entity, whose positions are same. Following the procedure for writing constraint relation, we select the fixed end of the string as the reference point.

Figure 8: Positions of moving parts from a fixed point.

Constraint relation

From the figure,

x A + x A + x B + x A + x B = L x A + x A + x B + x A + x B = L

⇒ 3 x A + 2 x B = 0 ⇒ 3 x A + 2 x B = 0

Differentiating w.r.t time, we have :

⇒ 3 đ x A đ t + 2 đ x B đ t = 0 ⇒ 3 đ x A đ t + 2 đ x B đ t = 0

There is an important subtle point here. The positions of blocks are on either side of the reference point (not on the same side as usually is the case). If positive direction of reference x-direction is towards right as shown in the figure, then velocities of two blocks are :

v A = đ x A đ t v A = đ x A đ t

v B = − đ x B đ t v B = − đ x B đ t

Substituting in the equation, we have :

⇒ 3 v A − 2 v B = 0 ⇒ 3 v A − 2 v B = 0

⇒ v B = 3 v A 2 ⇒ v B = 3 v A 2

The sign of this relation (positive) denotes that velocities of blocks "A" and "B" are both in the reference direction of x-axis.

Relative velocity of “B” with respect to “A” is :

v B A = v B − v A = 3 v A 2 − v A v B A = v B − v A = 3 v A 2 − v A

⇒ v B A = v A 2 = 4 2 = 2 m / s towards right ⇒ v B A = v A 2 = 4 2 = 2 m / s towards right

Problem 5 : A ball and rod of mass “m” and “nm” respectively are held in positions as shown in the figure. The friction is negligible everywhere and mass of the pulleys and the strings are also negligible. At a certain time, the ball and rod are released simultaneously. If the length of the rod is “L”, what time would the ball take to reach the other end of the rod. Consider "n > 1".

Figure 9: A ball travels past a vertical rod.

Vertical pulley system

Solution : The ball and the moving pulley are connected through the same string that goes over a fixed pulley. As such, their accelerations are same. Now, the rod is hanging from a string that passes over the moving pulley and its other end is attached to a fixed point. From the constrain relation as worked out in the subject module Pulleys, we know that acceleration of the rod is twice that of the moving pulley. Hence, we can conclude that acceleration of the rod is twice as that of the ball. Let us, then, consider that ball and rod move with accelerations “a” and “2a” respectively.

As the rod moves down (n >1), the ball moves up. The motions are in opposite direction. Since two elements are moving in opposite directions, the relative acceleration is arithmetic sum “a + 2a = 3a”. Now, using equation of motion for constant acceleration, we have :

L = 1 2 3 a t 2 L = 1 2 3 a t 2

⇒ T = ( 2 L 3 a ) ⇒ T = ( 2 L 3 a )

In order to evaluate this expression, we need to know acceleration “a”. From the free body diagram of the moving pulley, we note that the tension in the string attached to ball is twice the tension in the string attached to the rod.

Figure 10: Forces on the elements of the system.

Vertical pulley system

The free body diagram of the ball and the rod are shown on the right side of the above figure. The force analysis in the vertical directions yields following relations :

n m g − T = n m X 2 a n m g − T = n m X 2 a

2 T − m g = m a 2 T − m g = m a

Multiplying first equation by “2” and eliminating “T”, we have :

⇒ 2 n m g − m g = 4 n m a + m a = 4 n + 1 m a ⇒ 2 n m g − m g = 4 n m a + m a = 4 n + 1 m a

a = 2 n − 1 g 4 n + 1 a = 2 n − 1 g 4 n + 1

Putting this value in the expression of time, we have :

t = ( 2 L 4 n + 1 3 2 n − 1 g ) t = ( 2 L 4 n + 1 3 2 n − 1 g )