# Moment of load on arm.

SteelDirigibl

## Homework Equations

So moment and torque are really the same, right? so the force times the distance.

## The Attempt at a Solution

The force used has to be perpendicular to the line of the radius about O, so I assume I must do cos(36)*6lb*5in for the first. The second (the ball) confused me because it's not on the same line, so something must be missing right? or does it not matter?
I see the 54 deg. at the top, but its a little unclear, i assume between the arm and y axis. this is just for center of gravity of arm though, and would need a different angle to find the balls distance from O? or can we pretend it is the same?
I did, found the distance to be 8.899, and for the force it is cos(36)*9*8.899.

Then do you just add the two moments? I got 64.8 for the ball and 24.27 for the arm. added up these are 89.1 lb-in. is this right? and also I guess that should be negative because it is clockwise?

and obviously i can't do the second part until i finish the first, but i'm pretty sure i know how to do that.

Homework Helper
Gold Member
The moment of a force is the force times the perpendicular distance from the line of action of the force to the pivot. You didn't crunch out the numbers corrrectly when you found the ball's moment, because you went through an unnecessary step and made an error. What is the perpendicular distance from the line of action of the ball's weight force to the pivot? Otherwise, you are proceeding correctly.

Homework Helper
Hi SteelDirigibl!

(have a degree: ° )
So moment and torque are really the same, right? so the force times the distance.

Right.
The force used has to be perpendicular to the line of the radius about O, so I assume I must do cos(36)*6lb*5in for the first.

Correct.
The second (the ball) confused me because it's not on the same line, so something must be missing right? or does it not matter?

Only the distance to the line of application of the force matters (and you're given that!).

The distance to the point of application of the force doesn't matter.

SteelDirigibl
so I just use 11" as the distance? 99 lb-in for the ball?
but why isn't that the case for the weight of arm? the distance to point of application of the arm was given and was used, but that isn't the distance to the line of application, is it? I got it right but now I'm a little more confused than before.

Homework Helper
Gold Member
so I just use 11" as the distance? 99 lb-in for the ball?
yes
but why isn't that the case for the weight of arm? the distance to point of application of the arm was given and was used, but that isn't the distance to the line of application, is it? I got it right but now I'm a little more confused than before.
There are 3 ways to determine the torque from the arm's weight.

The first is to find the perpendicular distance from the line of action of the force, and multiply it by the force. This distance is 5cos36,and the torque is 30cos36.

The 2nd way is to break up the arm weight force into its x and y components, then find the torque from each component and add them. In which case you get (6cos36)(5) + (6sin36)0 = 30 cos 36.

The 3rd way is to use the cross product rule Torque = r X F = rFsintheta, where r is the magnitude of the position vector between the pivot and the point of application of the load, and theta is the included angle in between the force and position vectors. In which case, Torque = (5)(6)sin126 = 30 cos36.

You get the same result..which method you choose depends on which is easier in your particular problem.

SteelDirigibl
ok. so i used the 3rd method for the arm, and the first for the ball then.

Homework Helper
oops!

Hi PhanthomJay!

I'm sorry, I didn't notice your earlier post when I made mine.