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

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In summary, the question is asking for the net torque about the axle, given a square with a side length of 0.1m and a pivot placed on the bottom left corner. Two equal force vectors of 50N each point parallel to the left and north sides of the square. The attempt at a solution results in a torque of 5Nm, but the provided answer is 4.3 Nm. There is a possibility of a typo in the solution, as it does not account for the gravitational force acting on the square. Working backwards, the mass of the square can be calculated as 14/g, which may seem odd, but can be attributed to a possible typo in the solution.
  • #1
blintaro
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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.
 

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  • #2
How do you know that anwer ?
 
  • #3
Looks like an incorrect answer was provided, or perhaps the square has a mass that's not mentioned in the problem statement?
 
  • #4
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
blintaro said:
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.
 
  • #6
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.
 
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Related to Net Torque Calculation for a Square with Equal Forces Acting on Opposite Sides

1. What is an odd torque problem?

An odd torque problem refers to a situation where the magnitude and/or direction of the torque applied to an object does not result in the expected rotational motion.

2. What causes an odd torque problem?

There can be a variety of factors that contribute to an odd torque problem, such as imbalances in the forces acting on the object, friction or other external forces, or inconsistencies in the measurement or application of torque.

3. How can an odd torque problem be solved?

To solve an odd torque problem, it is important to first identify the source of the issue. This may involve examining the forces acting on the object, rechecking measurements and calculations, or adjusting the application of torque. Once the source of the problem is identified, steps can be taken to address it and achieve the desired rotational motion.

4. Are odd torque problems common in scientific experiments?

Odd torque problems can occur in any situation where torque is applied to an object, including in scientific experiments. However, with careful planning, consistent measurements, and proper application of torque, these issues can often be avoided.

5. How can odd torque problems be prevented?

To prevent odd torque problems, it is important to carefully plan and design experiments, accurately measure and record data, and carefully apply torque using appropriate tools and techniques. It can also be helpful to have a thorough understanding of the forces and mechanics involved in the experiment to anticipate and address any potential issues.

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