Finding acceleration of pushed crate

  • Thread starter Thread starter missashley
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The problem involves calculating the acceleration of a 900 N crate pushed with a 313 N force at a 25-degree angle, with a coefficient of kinetic friction of 0.17. The mass of the crate is determined to be approximately 91.743 kg. The frictional force opposing the motion is calculated to be 175.4875 N. After accounting for the net force in the x-direction, the correct acceleration is found to be approximately 1.179 m/s². The discussion highlights the importance of correctly incorporating friction and performing accurate arithmetic to arrive at the final answer.
missashley
Messages
34
Reaction score
0

Homework Statement


A 900 N crate is being pushed across a level floor by a force of 313 N at an angle of 25 degree above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.17. Acceleration of gravity is 9.81 m/s^2

http://img233.imageshack.us/img233/3717/31nk3.jpg

What is the acceleration of the box in m/s^2?

Homework Equations



F = ma

The Attempt at a Solution



Fg = 900 N
mass of crate = 91.743 kg
U = 0.17
g = 9.81 m/s^2
theta = 25
Fn = 900 + 132.2795
Ff = 0.17 * Fn = 175.4875

313cos 25 = 283.674 N
313sin 25 = 132.2795 N

Fx = ma
283.67 = 987.743a
a = 3.201 m/s^2

I didn't know whether or not to incorporate the y direction forces
 
Last edited by a moderator:
Physics news on Phys.org
Looks like you forgot to include friction.
 
Is the box moving up or down? What would this infer about acceleration?
 
Gannon said:
Is the box moving up or down? What would this infer about acceleration?

The box is moving left not up or down. Since the force is applied at an angle, the acceleration is not as great if applied directly horizontal
 
Doc Al said:
Looks like you forgot to include friction.

F = UFN
Friction = 0.17* (900 + 132.2795)
Friction = 175.4875 N
 
missashley said:
F = UFN
Friction = 0.17* (900 + 132.2795)
Friction = 175.4875 N
Good. Now recalculate the total force in the x-direction. (Which way does friction act?)
 
Doc Al said:
Good. Now recalculate the total force in the x-direction. (Which way does friction act?)

Friction act against the block, therefore, total force in x direction:

283.67 - 175.4875 = ma
111.1825 = 91.7431a
111.1825 / 91.7431 = a
a = 1.211889 m/s^2

did I do that right?
 
Looks good. (Round off your answer to a reasonable number of significant figures.)
 
I plugged it into the site where we're suppose to turn in our answers and it says it wrong. Hmmm
 
  • #10
arithmetic error!

missashley said:
283.67 - 175.4875 = ma
111.1825 = 91.7431a
Oops... double check that subtration. (Sorry for not catching that earlier.)
 
  • #11
yay it worked thanks so much! The answer is 1.179 m/s^2
 

Similar threads

Back
Top