Question

A 50-kg, 10-m-long lever has a fixed pivot point at one end and is kept horizontal by a vertical cable attached at the other end of the lever. A 5-kg block is pushed from the pivot point towards there other end giving it a constant speed of 0.2 m/s. Determine the rate of change in the tension of the cable.

a. 0.9800 N/s

b. 9.8000 N/s

c. -9.8000 N/s

d. -0.9800 N/s

Answer 1

Magnitude of torque is given by: r*F, where r is distance between point of application of force and the point where torque is to be calculated.

In the given problem, if we consider torque about the end point of lever, then, r is increasing at the rate of 0.2 m/s. So,dr/dt=+0.2 m/s

Also,F=mg, whre m is mass of the block and g is gravitational acceleration. Here,m=5 kg. So, F=5*9.8=49 N.

Now, rate of change of torque=d(r*F)/dt=F dr/dt( as F is constant=49 N) = 49*0.2=9.8 N/s.

So,option B is correct.