Recent content by mx2ko

the length of the crate is .9 m and the height is .4 m. I don't understand what you mean by the lever arm of the tension is expressed in terms of the length of the crate and the given angles. Also, since the height is given, what would the lever arm for the weight be.

actually I don't even think the lever arm of the weight is half the length x cos24degrees, but i could be wrong. so in that case I don't know how to find either of the lever arms

so then there would only be the weight and the tension that have torques. the weights lever arm is half the length x cosine24degrees i think. but what would the lever arm for the tension be since that would be needed to find the tension

is the good place placing it on the spot where the crate touches the ground so the normal force and the friction are both zero? and its a typical axis not one that is tilted on an angle

I know that the net force in the x and y direction both equal zero and the net torque equals zero because the object is in equilibrium since he is pulling the crate at a constant velocity, which means there is no acceleration. I tried to solve for the tension force by making the net torque and...

A man drags a 67.0-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 24.0° above the horizontal, and the strap is inclined 64.0° above the horizontal. The center of gravity of the crate coincides with its geometrical...