Net Torque Calculation for a Square with Equal Forces Acting on Opposite Sides

[blintaro]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

Torque = (50N)(.1 m)= 5Nm

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

 

[BvU]

How do you know that anwer ?

 

[gneill]

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

 

[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.

 

[gneill]

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.

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.

 

[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.