Question

Two blocks of 7 kg and 5 kg are connected by a heavy rope of mass 4 kg. An upward force of 200N is applied as shown in the diagram. The tension at the midpoint of heavy rope is:

Answer 1

Hint: Remember force is a vector quantity and so its direction is also to be taken into account. We need to find the total forces acting on the system and then find the net force in order to find the acceleration of the system.

Complete step by step answer:

Let us take the value of g = 10$$m/s^2$$

The blocks and the rope both have mass. So, total mass of the system is 7+4+5= 16 kg

Force of gravity acts on the system and it always acts downwards, so the force is $$16g=160N$$

Tension is 200 N and it acts in the upward direction, since upward force is greater in magnitude than the downward force, thus the system moves upwards. Applying Newton’s second law here,

$$F=ma$$

$$200-160=16a$$

$$a=2.5m/s^2$$

So, the value of acceleration is 2.5 $$m/s^2$$and is moving upwards.

Now to find the tension at the midpoint of the heavy rope, the mass is uniformly distributed of the rope so at midpoint, 2 kg is upside and 2 kg is downside.

$$T=(7+2)(g+a)$$

$$T=9\times 12.5$$

$$T=112.5N$$

Thus, the tension at the midpoint of heavy rope is 112.5 N

Note: Such problems can be briskly solved using Newton’s equations of motions. we are given the question that the rope is homogeneous which means mass is uniform throughout. Also, we have taken the value of g = 10$$m/s^2$$.