Calculating Force at an angle with kinetic friction coefficient

[skibum143]

Homework Statement

A force, F, applied to a 12.1 kg crate at an angle (theta) of 41.0 degrees makes the crate move horizontally with a constant acceleration of 1.85 m/s^2. The coefficient of kinetic friction between the crate and the surface is uk=0.35. Calculate the magnitude of F.

Homework Equations

F = ma
Fx = F + friction + N + W = ma
Fy = F + friction + N + W = 0
uk*N = friction
so, F = mg(cos theta) - friction = ma

The Attempt at a Solution

I first found the force of kinetic friction, by multiplying the coefficient of kinetic friction (0.35) by the normal force (12.1 * 9.8), which equaled 41.5 N.
Then, I found m*a = (12.1 * 1.85) = 22.385.
I then set up the equation: F cos(theta) - friction = ma, or F(.7547) - 41.5 = 22.385
This gave me F = 84.65 N, which is wrong.
Can someone help me see where I'm going wrong?
Thanks!

 

[thebigstar25]

im not sure if i understand the question u have .. is the crate in a horizontal surface and the crate itself lifted to be at an angle 41 .. or the crate is moving in a plane which is at angle of 41 ?? and mention the direction as well is it upward or downward??

 

[skibum143]

Sorry, the crate is moving horizontally to the right, being pulled by a force F at an angle 41 degrees above the horizontal axis. (most likely a rope or something!)

 

[skibum143]

The crate is not moving up an angle. it is moving across a flat surface, but being pulled by a rope at a 41 degree angle. Is there any way to paste diagrams in here? I could try to scan it...

 

[thebigstar25]

I think that the setup of your problem is something like the following:


http://img52.imageshack.us/img52/8946/ed96dcopycopy.jpg



so, in that case u have calculated the friction force incorrectly ..

it is true that the friction = (uk) * normal force .. in your case in order to find the normal force you should use Newton`s 2nd law which is :

Fy(net force) = 0 , in your problem this is , normal force + Fy - weight = 0 .. normal force is not always mg as you suggested .. go on from this point and you should get the right answer ..

 

[skibum143]

But if Fy = 0, and normal force + Fy (0) - weight (mg) = 0,
then normal force = weight, so normal force = mg.

Am I missing something??

 

[skibum143]

Oh, but you are saying that:
normal + Fy (Fsin(41)) - weight (mg) = 0
So normal + Fsin(41) = weight
Can I use mg for the F of Fsin(41)?
That would give me a normal force of 40.785
and when I plug it into Fcos(41) - friction (40.785*0.35) = ma
I get F to be 48.568
Is this correct? Because I used mg for the F in the Fsin(41) but then I solved for the F in the Fcos(41)...I'm not sure that is right?

 

[thebigstar25]

noooooo you can't substitute mg for F .. Your questions is asking for the value of F ..

I suggest you solve the following equations :

Fx = Fcos41 - friction = ma
Fy = Fsin41+normal force-mg = 0
When you substitute you have friction = (ku) * normal ..

I think this is clear enough .. Try again and tell me what you did

 

[skibum143]

sorry, I'm horrible at physics. here is what I think you are trying to explain:

if I solve for normal force using Fy, I get:
normal force = mg - Fsin41
if I plug that into the Fx equation, I get:
Fx = Fcos41 - (ku)*(mg - Fsin41) = ma
If I solve for F, I get F = 122 N

Am I going the right way now?
Thanks for your help.

 

[thebigstar25]

what you are doing is right .. But i have solved your problem and the answer is not 122 ? .. I solved it twice so i don't give you a wrong answer .. You probably made simple mistake in your calculations .. Try again you should get some value between 60 and 68 ..

 

[skibum143]

I am so confused.
When I do the equation, I get:
Fcos41 - (ku)*(mg - Fsin41) = ma
.7547F - .35(118.58 - .6561F) = 22.385
.7547F - .2296F = 22.385 + 41.503
F = 121.67

Am I doing some trig function wrong here?

 

[thebigstar25]

ooooh u had made a simple mistake .. U have a sign mistake .. Ur equation should be :
0.7547F + 0.2296F = 22.385 + 41.503