Question

The figure below shows two blocks connected by a string of negligible mass passing over a frictionless pulley. m1 = 7.0 kg and θ = 12.0°. Assume that the incline is smooth. (a) For what value of m2 the will the system be in equilibrium? m2 = kg (b) If the block has to slide down the incline with an acceleration of 0.4 m/s2, what should be the value of m2? m2 = kg

Answer 1

First we assume there is no fiction, this is just a basic assumption of all physics problems. We also assume the rope is massless and does not change length since this problem is significantly more complicated if the rope has a mass or can change length.

For part (a) it asks about a system in equilibrium so we see that m1 is hanging vertically and if the system is in equilibrium then all the forces are constant. So I m1 is 7.0 kg then the force of gravity is

Fg = (7.0 kg) (10 N/kg) = 70 N

Since the forces are balanced there must be 70 N pulling it up. All that force is done by the other mass and transfered through the massless rope.

In this case the force of gravity on m2 is 336.681N (see the attached picture for explanation).

So if the force on m2 is about 337 N then it's mass is

m = (337 N) / (10 N/kg) = 33.7 kg

For part (b) it asks about a system where there are unbalanced forces. If m2 slides down with any acceleration and the rope cannot change length then m1 must be accelerating at the same rate. So if m1 accelerates at .4 m/s^2 then the force is

F = Fg + Fm2
F = 70 N + (7.0 kg) (.4m/s^2) = 70 N + 2.8 N = 72.8 N

So now the process for finding the mass of the other block is the same except instead of using 70 in the equation you use 72.8 N. So the new force is 350.149. So the mass must be equal to

m = (350) / (10 N/kg) = 35 kg