Maximum Pulling Force for Two Blocks and a Cord

In summary, the conversation discusses a problem involving two blocks of different masses connected by a string and being pulled horizontally. The question is to determine the maximum force that can be exerted on the system without causing the top block to slip, given the coefficient of static friction. The conversation includes a free body diagram and equations to solve the problem.
  • #1
mizzy
217
0

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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  • #2
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.
 
  • #3
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!
 

FAQ: Maximum Pulling Force for Two Blocks and a Cord

What is the concept of "Two blocks and a cord"?

The concept of "Two blocks and a cord" is a common physics problem that involves two blocks connected by a cord or rope that passes over a pulley. This setup is used to demonstrate concepts such as friction, tension, and acceleration.

How do you calculate the acceleration in "Two blocks and a cord"?

To calculate the acceleration in "Two blocks and a cord", you can use Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration. You can also use the equations of motion, which take into account the initial velocity, final velocity, acceleration, and time.

What is the role of friction in "Two blocks and a cord"?

Friction plays a crucial role in "Two blocks and a cord" as it affects the tension in the cord and the acceleration of the blocks. The force of friction opposes the motion of the blocks and can cause a difference in the tension on either side of the pulley.

How does the mass of the blocks affect the system in "Two blocks and a cord"?

The mass of the blocks affects the acceleration of the system in "Two blocks and a cord". The greater the mass of the blocks, the greater the force needed to accelerate them. This means that the acceleration will be smaller if the blocks have a larger mass.

What is the significance of the pulley in "Two blocks and a cord"?

The pulley in "Two blocks and a cord" serves as a point of support for the cord and allows for the transfer of tension from one block to the other. It also changes the direction of the force applied by the cord, which can affect the acceleration and tension in the system.

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