Physics question......?

Question

I'm stuck on a Physics problem-
A 7.5 kg block is resting on a horizontal surface, and is connected to a 5.50 kg mass suspended from a string passes over a frictionless pulley. Find the acceleration of the system and the tension in the string if the surface is:
(a)
smooth, and
(b)
the coefficient of friction is 0.190

I have found the acceleration of the system with the smooth surface, but i can't find the tension in the two strings or the acceleration of the system with the coefficient of friction.....

Any help is greatly appreciated.....

Answer 1

mavs is right.
karlterzaghi didn't do the problem right because he ignored the fact that part of the force of gravity (on the 5.5kg mass) is used to accelerate the 5.5kg mass. Only the remainder of the force is available to produce tension in the string and to accelerate the 7.5kg mass.

The simple way to think of it is that you have a force of 5.5 x 9.81N to accelerate a total mass of 13kg in the no-friction example. And you have a force of 5.5 x 9.81N - 7.5 x 9.81 x .190N in the other example, again accelerating a total mass of 13kg.

In each case, you can solve for the acceleration. Then use the acceleration value to solve for the force needed to accelerate the 7.5kg mass at that rate (and to overcome friction in the second part). That force equals the tension in the string. [Or you could solve for the force restraining the 5.5kg mass (i.e., reducing its acceleration below that of freefall) in order to find the tension in the string.]

If you take that approach, you should come to the same result as with mavs's method.

Answer 2

Since there is no friction at the contact between the string and the pulley (I assume the coeff. of friction only applies at the contact between the 7.5 kg block and the horizontal surface), in both cases the tension in the string will be equal to 5.50kg*9.81m/s^2 = 53.96 Newtons.

Now, in both cases you need to apply F=m*a to find the acceleration of the 7.5 kg block.

For the first case F = 53.96 N, so a = F/m = 53.96 N/7.5 kg = 7.19 m/s^2 (or 0.73 times the acceleration of gravity, 0.73g).

For the second case F = 53.96 N - 0.190*7.5*9.81 N = 39.98 N, so a = F/m = 39.98 N/7.5 kg = 5.33 m/s^2 (or 0.54 times the acceleration of gravity, 0.54g).

I hope this helps.

Answer 3

Set up equations for both masses:

for the 5.5 kg block, mg - T = ma
for the 7.5 kg block, T - c Mg = Ma

where, c is the coefficient of friction. If c =0, then the equation just becomes T = Ma

You can solve for T and a using this 2 equations and substituting for masses and g=9.8 m/s^2.