# Confusion about how to identify lever arm

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1. Dec 28, 2015

### Chozen Juan

1. The problem statement, all variables and given/known data
A rotational axis is directed perpendicular to the plane of a square and is located as shown in the drawing. Two forces, F1 and F2, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the drawing. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of F2 is three times that of F1. Find the distances a and b that locate the axis.

2. Relevant equations
Torque
= r x F
|Torque| = |r| |F| sin(angle)

3. The attempt at a solution
I first tried to attempt this by setting up equations for the net torque of both cases. I realized that the net torque would have to be zero and I knew the forces, but I incorrectly thought that the distance from the force to the axis (or as the textbook likes to call it, the lever arm) is some hypotenuse of a triangle whose legs are somewhere along the sides of the square. As I tried to solve the problem and find the angle between the displacement and the force, I realized that I had too many unknown variables. I then looked at my solution guide, and it stated that the "lever arm"s are actually a and b themselves, and the angles between the forces and the lever arms are both just 90 degrees. With this information, I attempted the problem once again and solved it with ease:

(F2) =(3F1)
(F1)(b)-(F2)(a) = 0 -----> (F1)(b) = (F2)(a) ------> (F1)(b) = (3F1)(a) ------> b = 3a

(F1)(1-a) -(F2)(b) = 0 -----> (F1)(1-a) = (F2)(b) -----> (F1)(1-a) = (3F1)(b) -----> 1-a = 3b

System of equations:
1-a = 3b
b = 3a

1-a = 3(3a) -----> 1-a = 9a -----> 10a = 1 -----> a = 0.1m

b = 3(0.1) -----> b = 0.3m

Still, I am really really confused about why a and b are the distances between the force and the axis. The forces act at the corners, so shouldn't the distances between the force and the axis be the distance of a line from the corner to the axis?

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• ###### IMG_9948.jpg
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2. Dec 28, 2015

### BvU

Forces can be moved along their working lines without changing the physical situation: e.g. in picture (a), F2 is applied to the upper right, but it could be applied to the lower right just as well: that would give exactly the same situation.

Moving a force vector in a direction perpendicular to its working line is also possible, but then you have to add a torque to compensate: (I can't add a drawing right now, so it'll be in words): For example moving F2 on the upper right to the upper left can be done as follows: add a downward force with magnitude |F2| to the upper left plus an upward force with magnitude |F2| to the upper left. The downward one is F2 moved sideways and the upward one plus the original F2 are a torque and nothing else. Torque magnitude is 2|F2| x 1 m.  sorry, |F2| x 1 m

Does this make sense ?

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Last edited: Dec 28, 2015
3. Dec 28, 2015

### SteamKing

Staff Emeritus
It helps to remember that the moment of a force is defined as the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the axis of rotation:

Don't get distracted by the details of the object to which the force or forces are applied.

4. Dec 28, 2015

### Chozen Juan

I think I'm beginning to understand the part about how forces can be moved along their working lines, but I think I'd be able to understand the second situation with the force perpendicular to the working line better if I saw a diagram. I tried drawing what you said, but I just got two torques on the upper left corner with magnitude (F2)(1-a) that cancel each other.

5. Dec 28, 2015