# Statics Problem: Weight Hanging from a Rope

1. Jul 20, 2010

### wwshr87

1. The problem statement, all variables and given/known data
Mass hanging from the pulley in 50kg, the mass on the rope is 1kg. Please see attachment for details. The answer will be in terms on L.

2. Relevant equations
The sum of the forces in the y-direction and x-direction are equal to 0.

3. The attempt at a solution
Adding the moments at the pulley and equating them to 0; we have that T=50.
Now a free body diagram of the second weight. 2*50sin(x)=1, sinx=(1/100)=H/X
X=(L^2+H^2)^(1/2), substituding this in (1/100)=H/X, and solving for H, I find H=L/100.
Which is not the answer I have in the solutions.
Any help will be greatly appreciated.

#### Attached Files:

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• ###### staticproblem1.png
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2. Jul 20, 2010

### 6Stang7

Can you explain to me "2*50sin(x)=1"?

3. Jul 20, 2010

### wwshr87

When I draw the FBD of the 1kg weight; I addthe forces in the y-direction
50*sin(x)+50*sin(x)=1

4. Jul 20, 2010

### 6Stang7

What is the answer that they give?

5. Jul 20, 2010

### wwshr87

#### Attached Files:

• ###### solutionprob5.png
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Views:
128
6. Jul 20, 2010

### 6Stang7

The key here is that they are asking how much the 50kg weight will move up due to the 1kg weight. the Solution shows the change in length of the rope over the horizontal section, which is the same as the change in height of the vertical section, which is the same as the 50kg mass.

7. Jul 20, 2010

### wwshr87

Ok, I understand that now. But I still don't get where the 50/L is coming from.

8. Jul 20, 2010

### 6Stang7

sin(tetha)=50/L

9. Jul 20, 2010

### wwshr87

Ok got it, thanks for the explanation.