Calculating Coefficient of Friction

[listera]

Help me! (mu K)

Will someone PLEASE explain to me how to find the coefficient of friction when given the following:

a box weighing 325 N moving with constant velocity across a floor, and it is being pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below horizontal.

No matter how many times it is explained to me, I can't get the proper equation for Fn. Help! (test Monday)

 

[Brad Barker]

Will someone PLEASE explain to me how to find the coefficient of friction when given the following:

a box weighing 325 N moving with constant velocity across a floor, and it is being pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below horizontal.

No matter how many times it is explained to me, I can't get the proper equation for Fn. Help! (test Monday)

Ok, there is a constant velocity, so by Newton's First Law, there is no net force on it.

That means that, by Newton's Second Law, the two forces must be equal in magnitude in the direction of motion (let's call that the x-direction) and opposite in direction. The forces in the y-direction must be equal and opposite, too, as there is no acceleration in that direction as well.

Now, you are going to want to resolve that applied force vector first!

The magnitude of the force component in the x-direction is: 425 cos(35.2 deg).

The magnitude of the force component in the y-direction is: 425 sin(35.2 deg).

Now we're ready to set up two equations using Newton's Second Law...

F_x, net = 0 = 425 cos(35.2) - mu_k * N

F_y, net = 0 = N - mg - 425 sin(35.2)

Now just solve the second one for N (the normal force) and the first one for mu_k, then plug in for N.

 

[Doc Al]

Start by identifying all forces acting on the box. Then, since you know that the velocity is constant the net force on the box must be zero. So that means that:
(1) the sum of the x-components of the forces = 0
(2) the sum of the y-components of the forces = 0

Express that mathematically. You'll need to know about friction: the force of kinetic friction is given by $f = \mu N$, where N is the normal force. You should be able to solve for $\mu$.

 

[listera]

Thanks a bunch!
(I didn't know that mg was subtracted from Fn!)

 

[Gokul43201]

Thanks a bunch!
(I didn't know that mg was subtracted from Fn!)

Do you understand why it's subtracted ? This is important !

Also, I hope you're drawing free-body diagrams...

 

[listera]

muchos gracias! You guys helped a lot; my thanks to ya!

 

[listera]

actually, i don't really know why it's subtracted... I tried looking it up, but in vain!

 

[Gokul43201]

Have you drawn a free-body diagram ? What are all the forces in the y-direction ?

 

[listera]

yeah I drew a body diagram...the forces I have in the y direction are the natural force and gravitational force.

 

[Doc Al]

yeah I drew a body diagram...the forces I have in the y direction are the natural force and gravitational force.

What you call Fn (and I call N) stands for "normal" force, not natural force. "Normal" means perpendicular; the normal force between the floor and the box is the force that the floor exerts "normal" to the surfaces--that means straight up, in this case.

These are the forces acting on the box:
(1) weight = mg (acting down)
(2) N = normal force of the floor pushing up
(3) the applied force of 425 N (acting at a downward angle of 35.2 degrees below horizontal)
(4) the friction force of the floor against the box, f = $\mu N$ acting opposite to the direction of motion.

Note that the first three of these forces have components in the y direction.

 

[listera]

hey thanks...i finally figured it out. (looking back...I found myself in a 'duh' moment there...) lol