Maximum Pulling Force for Two Blocks and a Cord

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SUMMARY

The maximum pulling force that can be exerted on a system of two blocks, where a 5.00 kg block is on top of a 10.0 kg block, is determined by the coefficient of static friction, which is 0.350. The static frictional force (Ff1) acting on the top block can be calculated using the formula Ff1 = μ * m * g, where μ is the coefficient of static friction, m is the mass of the top block, and g is the acceleration due to gravity. The total frictional force (Ff2) acting on the bottom block is given by Ff2 = μ * (m + M) * g. By setting up the equations of motion and eliminating acceleration (a), the maximum pulling force (F) can be derived.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of static friction and its coefficient
  • Ability to draw free body diagrams (FBD)
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate the maximum static friction force for different coefficients of friction
  • Explore the effects of varying masses on the system's acceleration
  • Learn about dynamics of multiple body systems in physics
  • Study the principles of friction in inclined planes
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of static friction and free body diagrams in problem-solving scenarios.

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Homework Statement


A block of mass 5.00kg rides on top of a second block of mass 10.0kg. A person attaches a string to the top block and pulls the system horizontally. Friction between the two blocks keeps the 5.00kg block from slipping off. If the coefficient of static friction is 0.350, what is the maximum force that can be exerted by the string on the block without causing the top block to slip.


Homework Equations


F=ma

The Attempt at a Solution


First of all, I drew my FBD:

top block: gravitational force going down, normal force going up, Pulling Force going to the right and frictional force to the left

bottom block: gravitational force going down (mass of top and bottom block?), normal force going up. I have a feeling that I'm missing something in this diagram.

Can someone help me? thanks
 
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top block - mg downwards (m = 5), N2 upwards, F towards the right, Ff1 towards the left.

bottom block FBD - Mg downwards (M = 10), N1 upwards and friction (Ff1) towards the right on top, and Ff2 towards the left at the bottom.

And the whole system is accelerating towards the right with 'a', so you can the pseudo forces to the two blocks.

Note that Ff1 = [tex]\mu[/tex]mg ans Ff2 = [tex]\mu[/tex](m+M)g

You can now get two equations, eliminate 'a' and find F.
 
graphene said:
top block - mg downwards (m = 5), N2 upwards, F towards the right, Ff1 towards the left.

bottom block FBD - Mg downwards (M = 10), N1 upwards and friction (Ff1) towards the right on top, and Ff2 towards the left at the bottom.

And the whole system is accelerating towards the right with 'a', so you can the pseudo forces to the two blocks.

Note that Ff1 = [tex]\mu[/tex]mg ans Ff2 = [tex]\mu[/tex](m+M)g

You can now get two equations, eliminate 'a' and find F.



Thanks!
 

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