# Rotational Dynamics, Torque, and Tension

## Homework Statement

A man drags a 71.5-kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted 22.5° above the horizontal, and the strap is inclined 64.0° above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap. (Image attached)

torque = F * l

## The Attempt at a Solution

Since the crate is not rotating, the torques must balance. From what I can
see, there are 3 total torques - two clockwise and one counterclockwise.

I set up my equations like this:

t1-t3 are the 3 torques.
T is the tension on the rope.
theta1 is the angle between the box and horizontal.
theta2 is the angle between the rope and horizontal.

Clockwise Torques
t1 = mg * .9/2 * cos(theta1) <--- Torque due to gravity

t2 = T * cos(theta2) * .9 * sin(theta1) <- Torque due to horiz. comp. of tension

Counterclockwise Torque
t3 = T * sin(theta2) * .9 * cos(theta1) <- Torque due to vert. comp of tension

Then,

t1 + t2 = t3 ; Solved for T.

I have a feeling I'm setting up the torque due to gravity wrong.

Any thoughts?

#### Attachments

• 4.2 KB Views: 357

Related Introductory Physics Homework Help News on Phys.org
haruspex
Homework Helper
Gold Member
t1 = mg * .9/2 * cos(theta1) <--- Torque due to gravity
Imagine increasing the height of the box greatly. That wouldn't alter your equation above. Would it change the torque due to gravity? (Similar issue with the other torques.)

I'm thinking now that there is no torque due to gravity (no lever arm) and that the fact that the box is in equilibrium with a constant velocity has some importance.

That is, there is one equation for torque from tension: t1 = T*sin(theta2) * .45*cos(theta1)
and the other torque equation to cancel that one has to do with the normal force.

?

haruspex
Homework Helper
Gold Member
I'm thinking now that there is no torque due to gravity (no lever arm)
For torque, you have to consider the line through which a force acts and how close this gets to the point about which you are taking moments. Since the box is uniform, gravity through its centre. Does the line of its action pass through the point of contact of the box with the floor? If not, how much does it miss it by.
and that the fact that the box is in equilibrium with a constant velocity has some importance.
Not much. It means you're dealing with dynamic friction, not static, but there are no accelerations, so forces must still be in balance.

Torque as per wikipedia: is the tendency of a force to rotate an object about an axis,fulcrum, or pivot.
Choose which of these it is and choose your rotation point on the crate.

As haruspex stated, if one of the forces (gravity) passes through the point, it will does not create a torque around that point. Hence the question: Does the gravity force of the create pass through this ppoint? If not, what is the minimum distance between the pivot point and the imaginary infinite line representing the force(perpendicular distance).

mirahj: I think you have a good understanding of torque, just take a step back and look at the basics.