Calculating Force of Friction: A Homework Help Guide

[tyler hartman]

Force of friction help!

Homework Statement

A force of 755N is exerted on a rope to pull a 45.0N wooden crate across a concrete floor. If the rope makes an angle of 35.0° with the floor and the crate is moving at a constant velocity what is the amount of the force of friction?

Homework Equations

F=ma

The Attempt at a Solution

So far I have x: F=Fcosθ-Ff=ma I am not sure if that's the right equation given mg, Fa and angle

 

[haruspex]

Please don't use the same label (F in this case) for two different entities. Let's call the rope tension T. So, yes, you have F_x = Tcos(θ) - F_f = ma. What do you know about a?

 

[tyler hartman]

It equals 0 because its at a constant velocity? I think haha

 

[haruspex]

Quite so.

 

[tyler hartman]

So that means that Ff=Tcosθ?

 

[haruspex]

Yes. I assume there's more to this question. Do you have to find the coefficient?

 

[tyler hartman]

no, just the force. I am confused on how the weight of the box is not used whatsoever in the end equation?

 

[SammyS]

no, just the force. I am confused on how the weight of the box is not used whatsoever in the end equation?

The only horizontal forces are friction and the horizontal component of the applied force. Therefore, the weight of the box is not involved.

It is possible to find the coefficient of friction from the given information. To so this, the weight of the box would be involved.

 

[Basic_Physics]

Due to the weight of the box we get that the force of friction is a certain magnitude. From the fact that the box is not accelerating we then know that horizontal component of the pull must be equal to the force of friction and because we can calculate this from the information we have the force of friction. This is exactly how the force of friction is measured experimentally - pull the block along at a constant speed with a spring scale and take its reading. The force of friction is then eual to the reading on the spring scale if it was held horizontally during the pull.