# Homework Help: Odd torque problem

1. Aug 13, 2014

### blintaro

1. The problem statement, all variables and given/known data

The question asks, What is the net torque about the axle? Pictured is a square of sidelength .1 m, the pivot is placed on the bottom left corner. A force vector of magnitude 50 N points parallel to the left side and toward the pivot. Another equal force vector points north along the right side of the square. At the risk of being chastised I have included a picture.

2. Relevant equations
Torque = FxR or force(moment arm)
A force vector pointing toward the pivot exerts no torque.

3. The attempt at a solution
Torque = (50N)(.1 m)= 5Nm

Inexplicably (to me), the answer is 4.3 Nm.

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2. Aug 14, 2014

### BvU

How do you know that anwer ?

3. Aug 14, 2014

### Staff: Mentor

Looks like an incorrect answer was provided, or perhaps the square has a mass that's not mentioned in the problem statement?

4. Aug 14, 2014

### blintaro

4.3 Nm is the answer provided in the back of the book. There doesn't appear to be anything about the mass of the square, but even if there was, wouldn't the answer remain 5.0 Nm? I suspected the solution was wrong but I thought I'd post it to make sure I wasn't missing something.

5. Aug 14, 2014

### Staff: Mentor

A gravitational force would act vertically through the center of mass, which is not aligned with the point of rotation, but a typo in the solutions is the more likely scenario.

6. Aug 14, 2014

### blintaro

Oh yeah, that would make sense. I suppose one could work backward to find the mass of the square give the net torque. Would it be accurate to say if the torque from gravity is .7 Nm, then .7 = (mg)(.05) => m = 14/g?
Seems an odd number. Oh well, I think i's fair to chat this up to a typo in the solutions as well. Thanks for your time.

Last edited: Aug 14, 2014