Co-efficient of Friction (Limiting Friction)

  • Thread starter Thread starter FeDeX_LaTeX
  • Start date Start date
  • Tags Tags
    Friction
Click For Summary
SUMMARY

The problem involves calculating the coefficient of friction for a box with a mass of 40 kg being pulled by a rope at an angle of 33° with a tension of 190 N. The correct coefficient of friction, calculated using the formula Fmax = uR, is determined to be 0.55 when applying the equations of motion and resolving forces. The user reports discrepancies with the MyMaths website, indicating a potential error in their calculations or understanding of the problem. The calculations for the normal force (R) and maximum frictional force (Fmax) are accurately derived from the given parameters.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of trigonometric functions in physics
  • Familiarity with the concept of friction and its coefficient
  • Ability to resolve forces in two dimensions
NEXT STEPS
  • Review the principles of static friction and its calculation
  • Study the application of trigonometry in resolving forces
  • Learn about the dynamics of inclined planes and tension forces
  • Explore common errors in physics problem-solving and how to avoid them
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and friction, as well as educators looking to clarify concepts related to force resolution and friction calculations.

FeDeX_LaTeX
Science Advisor
Messages
436
Reaction score
13

Homework Statement


"A box of mass 40kg is pulled by a rope inclined at 33° to the horizontal. The box is about to slide and the tension in the rope is 190N. What is the co-efficient of friction (2 d.p.)?"

Homework Equations


Fmax = uR
g = 9.8 ms-2

The Attempt at a Solution


First, a diagram of the problem;

[PLAIN]http://img215.imageshack.us/img215/6131/m1stuck2.png

To find R, resolve upwards;

R + 190sin33° = 40*9.8
R = 392 - 190sin33°

Resolving horizontally to find Fmax;

Fmax = 190cos33°

So, the co-efficient of friction should be Fmax divided by R;

(190cos33°)/(392 - 190sin33°) = 0.552295128(...) = 0.55 (2 d.p.)

That is my total working for the problem, but the website I am using "MyMaths" is telling me it is incorrect, and I am assuming I must be incorrect because every time a new randomly-generated problem is created it tells me I am wrong when I solve that one using the method which I have described.

Could anyone please tell me the error in my method?

Thanks.
 
Last edited by a moderator:
Physics news on Phys.org
I also get .55
 

Similar threads

Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
22K
  • · Replies 13 ·
Replies
13
Views
8K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
31
Views
2K
Replies
5
Views
2K