# Homework Help: Friction / Tension problem

1. Oct 19, 2006

### Destrio

Hey,
I so far haven'y encountered a problem like this yet and dont know where to begin.

An object of mass, m , is suspended by two cords connected to a wall and to a 50 kg block resting on a table as shown. A coefficient of friction on 0.47 exists between the 5.0kg block and the table. What is the maximum mass, m , that can be hung from the cords before the 5.0kg block begins to move?

So far all I have been able to get is
Fnet = 0 due to no movement
and
Ff = uFn
= (0.47)(5.0kg)(9.8m/s^2)
= = 23.03 N

Where do I go from here:

Image: http://img85.imageshack.us/img85/3025/physicsfrictionib5.jpg [Broken]
http://img85.imageshack.us/img85/3025/physicsfrictionib5.jpg [Broken]

Last edited by a moderator: May 2, 2017
2. Oct 19, 2006

The first step is done - you calculated the friction force. Now, the maximal force in the horizontal string must equal the frictional force, right? So, you must set up the equations of equilibrium for the point (node) where the three strings meet.

3. Oct 19, 2006

### Staff: Mentor

Analyze the forces acting at the connection point. Vertical and horizontal forces must each add to zero. Hint: Call the (unknown) tension in the upper cord T and set up two equations.

4. Oct 19, 2006

### Destrio

Hey
So would the horizontal force on the right rope (Fx) be equal to the Ff , 23.03N?
and the Fy be 14.39N?

if so, how do I use this to determine the mass?

5. Oct 19, 2006

### gunblaze

Yes. The max Fx allowed will be the frictional force acting on the 5kg block. You now know that the tension of the string connected to the wall and connection pt does have a vertical pulling force and rightward pulling force. If the max allowable rightward pulling force is given by friction, are you able to find the upward pulling force by the tension of the string connected to the wall and connection pt? That upward force will be equal to ur max weight allowed by ur mass right?

6. Oct 19, 2006

### Destrio

I'm confused,
could you show an equation (or some of it) showing what you mean please?

7. Oct 20, 2006

As said above, set up the equation of equilibrium for horizontal compoments at the connection point first: $$-T_{1} + T_{2}\cos(180 - (90+23))=0$$, where T1 is the tension in the horizontal string (which equals the frictional force), and T2 is the tension is the string connected to the wall. Then, set up the equation of equilibrium for the vertical compoments: $$T_{2}\cos(23) - mg = 0$$. Since you calculated T2 in the previous step, I'm sure you see how to get the mass m now.

8. Oct 20, 2006

### Destrio

thanks!
I figured it out :)
I was thrown off for a moment when you put cos 23 (should be cos 32) but that helped me solve it
thanks very much

EDIT:
sorry, I didnt realize i wrote the wrong value on my diagram, my mistake

Last edited: Oct 20, 2006