Problem : A block of mass “m” is pulled by a string on a smooth horizontal surface with acceleration “a”. The string maintains an angle “θ” with horizontal. Find (i) tension in the string (ii) force applied by the surface on the block and (iii) force applied by the block on the surface.

Figure 1

Unbalanced force system

Solution : Here, we consider block as the body system.

Free body diagram of the block Free body diagram of the block

The external forces on the block are (i) Tension in the string,T, (ii) weight of the block, mg, and (ii) Normal force applied by the surface,N.

Figure 2

Free body diagram

The block moves in x-direction with acceleration “a’. On the other hand, there is no motion involved in vertical i.e. y-direction as implied by the question. The forces in y-direction, therefore, constitute a balanced force system. Now,

∑ F x = T cos θ = m a x ⇒ T cos θ = m a ⇒ T = m a cos θ ∑ F x = T cos θ = m a x ⇒ T cos θ = m a ⇒ T = m a cos θ

and

∑ F y = N + T sin θ - m g = 0 ⇒ N = m g - T sin θ ∑ F y = N + T sin θ - m g = 0 ⇒ N = m g - T sin θ

Putting values of tension, T, in the above equation,

⇒ N = m g - m a sin θ cos θ ⇒ N = m g - m a tan θ ⇒ N = m g - m a sin θ cos θ ⇒ N = m g - m a tan θ

According to Newton’s third law, the surface is acted upon by the same magnitude of force that the surface applies on the block. Thus, force on surface is also N, acting downward.

Problem : Two blocks A and B of masses 10 kg and 20 kg respectively, connected by a string, are placed on a smooth surface. The blocks are pulled by a horizontal force of 300 N as shown in the figure. Find (i) acceleration of the blocks (ii) tension in the string and determine (iii) whether magnitudes of acceleration and tension change when force is applied on other mass? Consider, g = 10 m / s 2 g = 10 m / s 2 .

Figure 3

Unbalanced force system

Solution : Let us first investigate acceleration and tension for the given configuration in the figure above. Here, we see that two masses have same acceleration. We can use this opportunity to apply the concept of composite body system (comprising of both blocks). We neglect tension in the string as it constitutes internal force for the whole composite body system.

Let “a” be the acceleration in the direction of force. The mass of the composite body system, m c m c , is :

m c = 10 + 20 = 30 kg m c = 10 + 20 = 30 kg

Figure 4

Composite body system

Free body diagram of the composite body system Free body diagram of the composite body system

Figure 5

Free body diagram

∑ F x = F = m a x ⇒ 300 = 30 a ⇒ a = 10 m / s 2 ∑ F x = F = m a x ⇒ 300 = 30 a ⇒ a = 10 m / s 2

Note that this force equation does not depend on whether we apply force on A or B. In either case, the force equation remains same – only direction of acceleration changes with the change in the direction of applied force. It means that magnitude of acceleration of the two body system will remain same in either case.

In order to find the tension in the string, however, we need to consider individual block. Here, we consider block "A" for the simple reason that it is acted upon by a single force (T) in x-direction and analysis of force on block "A" will be simpler than that of block "B", which is acted by two forces (T and 300 N force).

Free body diagram of the body A Free body diagram of the body A

Figure 6

Free body diagram

∑ F x = T = m a x ⇒ T = 10 a ⇒ T = 10 x 10 = 100 N ∑ F x = T = m a x ⇒ T = 10 a ⇒ T = 10 x 10 = 100 N

Now, we answer the third part of the question : whether magnitudes of acceleration and tension change when force is applied on other mass? Let us consider the case when force is applied on body “A” as shown in the figure.

Figure 7

Free body diagram

Now, we consider force and acceleration of block "B" as it is now acted by only one force i.e. tension in the string in x-direction.

Free body diagram of the body B Free body diagram of the body B

Figure 8

Free body diagram

∑ F x = T = m a x ⇒ T = 20 a ⇒ T = 20 x 10 = 200 N ∑ F x = T = m a x ⇒ T = 20 a ⇒ T = 20 x 10 = 200 N

Thus, we see that acceleration (a) is independent, but tension (T) in the string is dependent on the point of application of the external force.

Problem : A block of mass “m” is pulled by a string on a smooth horizontal surface with force F = kt, where “k” is a constant. The string maintains an angle “θ” with horizontal. Find the time when block breaks off from the surface.

Figure 10

Unbalanced force system

Solution : Here, important thing is to know the meaning of "breaking off". It means that physical contact between two surfaces is broken off. In that condition, the normal force should disappear as there is no contact between block and surface.

Since normal force is directed vertically up, we need to analyze forces in y-direction only so that we could apply the condition corresponding to "breaking off".

Free body diagram of the block Free body diagram of the block

Figure 11

Free body diagram

∑ F y = N + k t sin θ - m g = 0 ⇒ N = m g - k t sin θ ∑ F y = N + k t sin θ - m g = 0 ⇒ N = m g - k t sin θ

Now, let t = t B t = t B (break off time) when N = 0 (breaking off condition)

⇒ 0 = m g - k t B sin θ ⇒ k t B sin θ = m g ⇒ t B = m g k sin θ ⇒ 0 = m g - k t B sin θ ⇒ k t B sin θ = m g ⇒ t B = m g k sin θ