Question

the velocity of A and B are shown in figure . find the speed (in m/s) of block C (Assuming that the pulley and string and ideal).​

Answer 1

Answer:

Speed 5 m/s.

Explanation:

If we let the length of the strings from the bottom be l1,l2,l3,l4 and they are constant.

Again, if block A is being moved by x2 and the block B is being moved by x1. So correspondingly the block c will slide down by x4 while l3 remain same.

So, (l1 - (x1 - x2)) + l2 - (x1 - x2) + l3 + (l4 + x4) = (l1 + l2 + l3 + l4).

After the (x1 - x2) change in position takes place so on solving x4 will come as 2(x1 - x2).

Now, on differentiating with respect to time we will get dv4/dt = 2(dx1/dt - dx2/dt) or v4= 2(v1-v2) so, v4=2(3-1)=4 m/s.

As we can see the block C is in contact with the block B so, net velocity of block C will be -3i - 4j or the speed which is the magnitude will be √(3^2+4^2) = 5 m/s.