Finding acceleration of a crate when applying a force at an angle

[Occidomortem]

Homework Statement

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate and ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate?

Force (horizontal)= 5000N
U_K= 0.22
m= 250kg
a (upward) = ?

Homework Equations

a=Fnet/m
F_k=U_kn

The Attempt at a Solution

The first thing I did was find the net force. In the upward direction the force is equal to cos(30)5000N, The force of gravity acting downward on the incline is sin(30)mg, the kinetic friction also pointing downward of the incline is F_k=.22n, where normal force is equal to cos(30)mg from the incline plus the force from applying the horizontal force sin(30)5000, n=cos(30)mg+sin(30)5000

So I got an upward force on the incline to be 4330.127N. and the downward for from gravity to be 1225N. The downward force from friction= 0.22(cos(30)mg+sin(30)5000) = 1016.79.

the Fnet upward is = 4330.127N-1225N-1016.79N = 2088.337

a=F/m

a=2088.337/250kg

a=8.35m/s^2, this however is wrong. I think I messed up with my normal force somewhere.

So I guess my main question is how does a horizontal force affect the normal force on an incline. I apologize for the bad formatting/english I'm not used to typing these equations ect.

 

[Doc Al]

a=8.35m/s^2, this however is wrong. I think I messed up with my normal force somewhere.

Why do you think it's wrong?

So I guess my main question is how does a horizontal force affect the normal force on an incline.

I think you calculated the normal force just fine.

 

[Occidomortem]

The answer should be 13m/s^2 for some reason. It is possible that my professor made a mistake.

 

[meztek]

The answer should be 13m/s^2 for some reason. It is possible that my professor made a mistake.

Your professor forgot to account for the force of gravity acting in the -X direction in the frame of the ramp. If you only consider friction to find the net force you end up with 13 m/s^2 (incorrect) instead of 8.2 m/s^2.

 

[kuruman]

Thank you for your contribution, however @Occidomortem is unlikely to profit from it because this thread is more than 9 years old and the OP hasn't visited PF since May 21, 2014.

 

[meztek]

Perhaps not, but someone else may profit from it when they come across this thread.

 

[gneill]

Hmm. Shouldn't the y-component of the Person's force be negative, given the direction of the y-axis?

 

[kuruman]

Perhaps not, but someone else may profit from it when they come across this thread.

Here is how to do it transparently and for the record.

Step 1
Draw a free body diagram.

Step 2
List all the entities that interact with the crate and write their components. I assume that the horizontal force is exerted by a person. I also orient the positive x-axis up the incline and the positive y-axis perpendicular and away from the incline.

Entity	x-component	y-component
Person	$F\cos\theta$	$\cancel{F\sin\theta}~-F\sin\theta$
Earth	$-mg\sin\theta$	$-mg\cos\theta$
Incline	$-\mu_k~N$	$N$

Step 3
Find the components of the net force and set each equal to mass times acceleration in that direction.
$F_{\text{net,x}}=F\cos\theta-mg\sin\theta-\mu_k~N=ma$
$F_{\text{net,y}}=-F\sin\theta-mg\cos\theta+N=0.$

Step 4
(a) Solve the second equation to find an expression for the normal force
$N=mg\cos\theta+F\sin\theta.$
(b) Substitute in the first equation and find an expression for the acceleration.
$F\cos\theta-mg\sin\theta-\mu_k~( mg\cos\theta+F\sin\theta )=ma$
$a=\dfrac{F}{m}(\cos\theta-\mu_k\sin\theta)-g(\sin\theta+\mu_k\cos\theta).$

Step 5
Substitute numbers and evaluate.
$a=\dfrac{5000~(\text{N})}{250~(\text{kg})}(0.866-0.22*0.5)-9.8~(\text{m/s}^2)(0.5-0.22*0.866)=8.35~(\text{m/s}^2).$

The professor's answer is incorrect and the OP is correct.
Note: This post has been edited in view of post #7 by @gneill. It seems that the professor and I made the same mistake.

 

[meztek]

Note: This post has been edited in view of post #7 by @gneill. It seems that the professor and I made the same mistake.

As I already said.