Is this correct?

Question

A 7.50 kg block is resting on a horizontal surface, and is connected to a 5.50 kg mass suspended from a string passed over a frictionless pully. The surface is smooth, with no friction.

Would the tension on the string be 53.9 N?

I'm also suppose to find the tension of the string if the coefficient of friction is 0.190 on the surface, I have no clue on how to find that..A little help maybe?

Answer 1

You have to consider that the weights are accelerating. Call the acceleration of the weights a (both are the same). The downward force on the suspended weight is m1*(g-a). The sideways force on the other weight is m2*a. The tension in the string is m1(g-a), and this is what is pulling the m2 weight, so its acceleration is given by m2*a = m1*(g-a), so a = (m1+m2)*g/m1. The string tension is m1*(g-a) = m1*[g - (m1+m2)*g/m1] = m1*g*[1-(m1+m2)/m1] = m1*g*m2/m1 = m2*g. When you add friction, the tension in the string is increased by the frictional force, which is m2*g*k, where k=coeff of friction. The total tension is then m2*g+m2*g*k, or m2*g*(1+k).

Answer 2

The coefficent of friction is м=f/r =0.190


In physics problems like this there CAN be a frictionless pulley even though it doesn't seem plausible.

In the problem you would have two different actions happening at once..the tension on each side of the pulley. So I set the problem up and do Sigma Fx and Sigma Fy. Then you use the equation from part two of the problem and plug it in to part one and get the tension of the string.

However, you stated that the tension on the string is 53.9N it correct. 5.50*9.81=53.9 or 54N.

Answer 3

The tension would be entirely due to the weight of the second mass, since the table is frictionless.

5.50 kg * 9.81 m/s^2 = 53.9 (well, actually 54.0) N

Your answer is correct.

Answer 4

Trick question. Their is no such thing as a frictionless pulley. There is always friction. It might be infinitesimally small, but there is always friction.