# Homework Help: Normal force and tension in a pulley system

1. Oct 29, 2009

### waldvocm

Block M(15.7kg) is on a horizontal table initially moving to the left because it is connected to a mass m(8.2kg) hanging vertically off of the edge of the table (left hand side of the table). A force F, with a magnitude of 75.3N, acts on M directed at an angle of 35.0 degrees above the horizontal( to the right hand side of the table). There is no friction and the pulley and string are massless.

a)What is the normal force (N) on M?

Fn=mg-75.3*sin(35.0)-T
Fn-15.7-43.19-52.79
Fn=-80.28??????????????

Or would I first subtract both of the forces acting on M T-F=52.79-43.19=9.6 Fn=15.7-9.6=6.1

b)What is the tension in the string

The tension origionally is 52.79 with out the force acting on M would it now be 9.6

c)What is the acceleration of M

d) Is the speed of M increasing or decreasing?

2. Oct 29, 2009

### kuruman

You do not show a diagram, but if I understand the problem correctly, the tension is a horizontal force while the normal force is in the vertical direcction. They do not belong in the same equation. Draw a free body diagram.

3. Nov 2, 2009

### waldvocm

Fn=mg-75.3*sin(35.0)
Fn=110.67

b) ax=m2g/m1+m2=8.2*9.80/15.7+8.2 a=3.36
T=m1*a T=52.79

c) how do I find the acceleration?

4. Nov 3, 2009

### kuruman

Your expression for the normal force is correct.

T = m1a is incorrect. The net force in the horizontal direction must include the horizontal component of force F which could be positive or negative depending on the direction of F. You will also need an equation for the hanging mass. This will give you two equations and two unknowns.

5. Nov 5, 2009

### waldvocm

do I use (may)=T cos(35.0)-mg and (max)=T sin(theta)

I am confused about sin and cos in the free body diagrams. It seems before that sin was used for y and cos for x. Why is it the other way around in the free body diagrams?

The Ts in both equations are equal to each other. So I am to somehow solve for T?