Horizontal Force on a Ramp required to Accelerate a Mass

Click For Summary
SUMMARY

The discussion focuses on calculating the horizontal force required to accelerate a 5.57 kg box up a ramp inclined at 38.0 degrees with a constant acceleration of 4.30 m/s². The coefficient of kinetic friction between the box and the ramp is 0.28. The initial calculation yielded a force of 23.951 N for acceleration and 12.044 N for friction, leading to a total horizontal force of 45.7 N. However, there are discrepancies in the calculations, particularly regarding the correct application of trigonometric functions and the force components involved.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Knowledge of trigonometric functions (sine and cosine) in physics
  • Familiarity with the concept of kinetic friction and its coefficient
  • Ability to resolve forces into components on an inclined plane
NEXT STEPS
  • Review the application of Newton's second law in inclined plane problems
  • Study the derivation of forces on an incline, including frictional forces
  • Learn how to properly use trigonometric functions in physics calculations
  • Practice similar problems involving horizontal forces and acceleration on ramps
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators looking for examples of force calculations involving friction and acceleration.

GiantCube
Messages
2
Reaction score
0

Homework Statement



A 5.57 kg box sits on a ramp that is inclined at 38.0^\circ above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.28.

What horizontal force is required to move the box up the incline with a constant acceleration of 4.30 m/s^2?

Homework Equations



Calculation F. F=ma is important. Trig functions. Calculation of F caused by friction.

The Attempt at a Solution



I calculated the force required to move the object up the ramp using a horizontal force; m*a/cos(38), which is 23.951N.

I calculated the force caused by the μk, which is sin 52*mg, which is 12.044 N

I added these, and found the horizontal component required, which is those added/ cos 28, which gives 45.7N

I am fairly certain I'm doing this wrong though, and I would appreciate any help.

Thanks!
 
Physics news on Phys.org
Welcome to PF, GiantCube.
I get the 12 and the 24 but not the 45.7. It would be nice to see your F = ma statement with the three things added to get the total force. We may have some differences in the signs.
 
Thanks. :)

The Force is the horizontal portion of the acceleration, so (m*a*cos(38)) must the force to get over the friction, so sin(52)*m*g. Which would be F= sin(52)*m*g + (u*m*a*cos(38))

... I'm fairly certain that's wrong, but not sure why...
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
15
Views
2K
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 56 ·
2
Replies
56
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
18
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K