pulley/tension prob!?

Question

A 15 kg block slides on a level frictionless surface while attached by a light string to a 5.0 kg hanging mass where the string passes over a massless frictionless pulley. What is the tension in the connenting pulley??

So confused, if you could help/explain I would appreciate it!!

thanks!!

Answer 1

In this type of problem, it is best to draw free body diagrams and look at the forces acting on each mass and sum since all forces will equal zero for static, and the net will be F=m*a for dynamic.

Also, the pulley has no friction or mass, so the tension in the string is equal on both blocks.

For this specific question, start with a look at the forces acting on the hanging mass:

I will use positive to indicate the downward direction

Gravity = m*g
The string tension -T
and the motion downward = m*a

so 5*a=5*9.81-T

Next, the forces acting on the sliding mass:
I will use positive to indicate the direction of motion
Since the surface is level, we only consider the forces parallel to the surface, so gravity is not a factor.
Next, the surface is frictionless, so there is no opposing force.
Finally, the only force acting on this block is the tension in the string T
so
m*a=T
or
15*a=T

Since the blocks are connected and there is no friction in the system nor inertia in the pulley, then the accelerations of both blocks are equal and the string tension is constant.

So we have simultaneous equations with two unknowns:
5*a=5*9.81-T
and
15*a=T

Since the question is asking for tension, I will solve for that variable first by isolating a:
a=9.81-T/5
a=T/15

subtract the first from the second:
0=T/5+T/15-9.81
solve for T
9.81=(3*T+T)/15
9.81*15/4=T
=36.8 N


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