# Torque a problem from Serway

## Homework Statement

Im having exam about Torque on monday and im trying to solve problems about it I stumbled with this problem, I cant solve it. I think im missing a fundamental concept about torque because of the fact that I dont understand how this problem was solved.

## The Attempt at a Solution

The problem is solved, I just need someone to tell me why when doing the sum of torques a 2000 cos 65 was used . I belive the torque should be the perpendicular component of the force to the rod or for the uniform boom. I wrote instead 2000 sin of 25 . any help please?

#### Attachments

please anyone? and why is the tension T divided as tx and ty in the sum of torques? T already has 90 degrees in respect to the axis , so shouldnt the torque of T just be T multiplied by its lenght?

PeterO
Homework Helper

## Homework Statement

Im having exam about Torque on monday and im trying to solve problems about it I stumbled with this problem, I cant solve it. I think im missing a fundamental concept about torque because of the fact that I dont understand how this problem was solved.

## The Attempt at a Solution

The problem is solved, I just need someone to tell me why when doing the sum of torques a 2000 cos 65 was used . I belive the torque should be the perpendicular component of the force to the rod or for the uniform boom. I wrote instead 2000 sin of 25 . any help please?

Torque = Force x Radius of operation

You either take the component of the force at right angles to the lever F.cos65
OR
You take the effective radius of operation R.cos65

SO either F x R.cos65 or F.cos65 x R giving F.R.Cos65

EDIT: they have done a vertical and horizontal for the second part

PhanthomJay
Homework Helper
Gold Member
please anyone? and why is the tension T divided as tx and ty in the sum of torques? T already has 90 degrees in respect to the axis , so shouldnt the torque of T just be T multiplied by its lenght?
Yes, you did find the torque of T the easy way...Serway first broke up T into its x and y components then summed torques of each.....to give the same result. But ultimately you need to break up T into its components anyway to get the reaction forces. Remember the 2 ways to find torques...Frsin theta or F times perpendicular distance from line of action of F to pivot point.

Also note that AB sin25 = AB cos 65 ....

Last edited:
Yes, you did find the torque of T the easy way...Serway first broke up T into its x and y components then summed torques of each.....to give the same result. But ultimately you need to break up T into its components anyway to get the reaction forces. Remember the 2 ways to find torques...Frsin theta or F times perpendicular distance from line of action of F to pivot point.

Also note that AB sin25 = AB cos 65 ....

The calculator wasent working well for some reason AB sin25 wasent the same as AB cos 65 i had to reset it. thanks a lot for your time .