How to find Friction Force and Coefficient of Friction?

Click For Summary
To find the friction force and coefficient of friction for a block sliding up a 20-degree incline, a horizontal force of 200 N is applied to a 15 kg block with an acceleration of 25 cm/s². The calculated friction force is 134 N, while the coefficient of friction is determined to be 0.65. The user struggles with the calculations, particularly in determining the normal force and its relationship to the friction force. Confusion arises from the discrepancy between the calculated coefficient of friction (0.97) and the expected value (0.65). Clarification is sought on the correct approach to finding the normal force and resolving the friction calculations.
softball1394
Messages
13
Reaction score
0

Homework Statement



A horizontal force of 200 n is required to cause a 15 kg block to slide up a 20degree incline with acceleration of 25 cm/s^2. Find the friction force on the block and the coefficient of friction.
ANSWERS:
Friction force = 134N
Coefficient of Friction = 0.65

Homework Equations


Ff = u Fn

The Attempt at a Solution


I can't figure out where to begin.
I'm not sure if it's relevant, but I found:
Fg to be 147.15 N
horizontal component to be 50.33 N
Vertical component to be 138.28N
 
Physics news on Phys.org
anyone there?

Part one:
Fgcos(20) - Ff = ma
138.28N - Ff = 15 x .25m/s^2
138.28N - Ff = 3.75
-138.28 -138.28
-Ff=-134.53
Ff=134.53N
(Hopefully this is correct?)

Part 2:
Ff = u Fn
u = Ff / Fn
u = 134.53N / 138.28N
u = .97

But the answer should be .65.
What am I doing wrong? The normal force should be 138.28 because 147.15cos(20) is 138.28, right?
Ugh, this is too confusing. Please help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
7
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
18
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K