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

• Occidomortem
In summary: The answer should be 13 m/s^2 for some reason. It is possible that my professor made a mistake.In summary, a 250-kg crate on a rough ramp, inclined at 30° above the horizontal, is being pushed up the ramp with a horizontal force of 5000 N. The coefficient of kinetic friction between the crate and ramp is 0.22. The acceleration of the crate is found to be 8.35 m/s^2, but the correct answer should be 13 m/s^2. This discrepancy may be due to a mistake made by the professor.
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) = ?

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.

Occidomortem said:
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.

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

Occidomortem said:
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.

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.

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

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

kuruman
meztek said:
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.

Last edited:
gneill and DeBangis21
kuruman said:
Note: This post has been edited in view of post #7 by @gneill. It seems that the professor and I made the same mistake.

1. What is acceleration?

Acceleration is the rate of change of velocity over time. It is a vector quantity that includes both magnitude and direction.

2. How do you calculate acceleration?

To calculate acceleration, you need to know the change in velocity and the time it took for that change to occur. The formula for acceleration is a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. How does applying a force at an angle affect acceleration?

When a force is applied at an angle, it can be broken down into its horizontal and vertical components. The horizontal component affects the acceleration in the horizontal direction, while the vertical component affects the acceleration in the vertical direction. This results in a change in the overall acceleration of the object.

4. How can you find the acceleration of a crate when applying a force at an angle?

To find the acceleration of a crate when applying a force at an angle, you can use the formula a = Fnet/m, where a is acceleration, Fnet is the net force acting on the crate, and m is the mass of the crate. The net force can be determined by considering the horizontal and vertical components of the applied force.

5. What units is acceleration measured in?

Acceleration is measured in meters per second squared (m/s²) in the metric system and feet per second squared (ft/s²) in the imperial system.

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