# Torque problem -- Squirrel standing on the end of a branch

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1. Nov 30, 2014

### Merrick

1. The problem statement, all variables and given/known data
A 600-g squirrel stands on the end of a 2.5-m-long branch that makes an angle of 60 degrees with the vertical trunk of the tree. What torque acts on the branch due to the squirrel?
2. Relevant equations

Torque= FxD torque=torque
3. The attempt at a solution
I'm pretty stumped on this one. I converted 600g to kg getting .6 kg. Then multiplied that by 9.8 to get a force of 5.88. Then I found the length of what I think is D by using sin(30) = x/2.5 to get 1.25. Then I did torque= 5.88(1.25) to get 7.35 Newton meters. I have a feeling i'm completely wrong though. Any help would be appreciated.

2. Nov 30, 2014

### Bystander

What makes you "think" this is D?

3. Nov 30, 2014

### Merrick

Well I thought getting a straight line would be the correct way of getting torque.

4. Nov 30, 2014

### Bystander

A straight line pointed what direction?

5. Nov 30, 2014

### Merrick

Horizontal to the ground

6. Nov 30, 2014

### Bystander

Good. And how do you get that from a 2.5 m branch which is 60 degrees to a vertical tree trunk that is perpendicular to the ground?

7. Nov 30, 2014

### Merrick

Cmon mate, throw me a bone here buddy. I just figured it made sense through a picture I drew.

8. Nov 30, 2014

### Bystander

How far is the horizontal distance from the end of the branch to the trunk of the tree? The branch is the hypotenuse of a rt. triangle, and you've been given two of the angles.

9. Dec 1, 2014

### CWatters

We can't see your picture but are you sure it's 2.5*sin(30)?

Sin = opposite/hypotenuse
so
Hypotenuse*sin = opposite