Two bloack stack on top, find max force to prevent skidding

  • Thread starter Thread starter vincer777
  • Start date Start date
  • Tags Tags
    Force Max
Click For Summary
SUMMARY

The discussion centers on calculating the maximum horizontal force that can be applied to block M without causing block m to skid off. The coefficients of static and kinetic friction between the blocks and the surface are denoted as mu-1, mu-2, mu-3, and mu-4. The derived formula for the maximum force is F(max) = (mu-4)(M + m)g - (mu-1)mg. The participant expresses uncertainty regarding the direction of static friction and its impact on the solution.

PREREQUISITES
  • Understanding of Newton's 2nd and 3rd laws of motion
  • Knowledge of static and kinetic friction coefficients
  • Ability to create and analyze free body diagrams (FBD)
  • Familiarity with basic mechanics concepts related to forces and motion
NEXT STEPS
  • Study the effects of varying coefficients of friction on force calculations
  • Learn how to construct and interpret free body diagrams for complex systems
  • Explore the implications of Newton's laws in multi-body systems
  • Investigate real-world applications of static and kinetic friction in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding frictional forces in multi-body systems.

vincer777
Messages
1
Reaction score
0

Homework Statement



A block of mass m rests on top of a block of mass M, which in turn rests on a horizontal surface. The coefficient of static friction between m and M is mu-1 and the coefficient of kinetic friction between m and M is mu-2. The coefficient of static friction between M and the surface is mu-3 and the coefficient of kinetic friction between M and the surface is mu-4. What is the maximum horizontal force you can apply to block M such that blocks m and M move at constant velocity and m does not skid off M.

Homework Equations



Newtons 2nd and 3rd law..


The Attempt at a Solution



Well, i solved the problem but i am still not convinced my answer is right.. I am assuming the problem i have is with Newtons 3rd law (deciding whether the static friection should point right or left). Here is my answer.

First, i made a free body diagram from the top block (m). The force down is mg, therefore the upward normal force is the force block M exerts on block m. I put the force of static fricition point left, because without fricition, nothing will keep the block moving to the left. Therefore, the force block M exerts on m (x-direction) is to the right.

Therefore,
F (M on m in y) = mg UP
F (M on m in x) = (mu-1)mg RIGHT


Now going to the FBD for the large box,

Obviously, F points left, force of kinetic friction points right, Mg points down, and n points up. Now using N3L,

F (m on M in y) = mg DOWN
F (m on M in x) = (mu-1)mg LEFT


Solving this i am left with

F(max) = (mu-4)(M + m)g - (mu-1)mg

But conceptually, i would think increasing mu-1 should increase the F max... so i doubt my answe is right.
 
Physics news on Phys.org
vincer777 said:

First, i made a free body diagram from the top block (m). The force down is mg, therefore the upward normal force is the force block M exerts on block m. I put the force of static fricition point left, because without fricition, nothing will keep the block moving to the left. Therefore, the force block M exerts on m (x-direction) is to the right.

You assumed the force F pointing to the left, did you?

The boxes interact through the static friction. If the force of static friction on m points to the left why is the force of the big box on m is to the right? The x-force of the big box is the same as the force of static friction.

ehild
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
13
Views
3K
Replies
4
Views
3K
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K