Newton's law problem: Pushing 2 stacked blocks on a horizontal table

In summary, the problem involves applying Newton's laws of motion to analyze the scenario of pushing two stacked blocks on a horizontal table. The key concepts include understanding the forces acting on both blocks, such as applied force, friction, and weight. By calculating the net force and acceleration, one can determine how the blocks move together as a single system or individually, depending on the frictional interaction between the blocks and the table. The problem illustrates principles of dynamics and the relationship between force, mass, and acceleration.
  • #1
JMAMA
18
0
Homework Statement
solve for acceleration of block A
Relevant Equations
f = ma
Would anyone be able to help with this Newtons law problem
Block A rests on top of block B as shown in (Figure 1). The table is frictionless but there is friction (a horizontal force) between blocks A and B. Block B has mass 6.00 kg and block A has mass 2.00 kg. If the horizontal pull applied to block B equals 12.0 N, then block B has an acceleration of 1.80 m/s2. What is the acceleration of block A

So my understanding is that from F = ma we can determine for block B that F = (6 kg)(1.80 ms2) = 10.80 N we can then say that block A is experiencing the same force yielding F - F(friction) = m a or 10.8 N - (2.0kg)(g)u = (2kg)(a). However i do not know how to solve for acceleration of A without the coefficient of friction
 

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  • #3
So my understanding is that from F = ma we can determine for block B that F = (6 kg)(1.80 ms2) = 10.80 N we can then say that block A is experiencing the same force yielding F - F(friction) = m a or 10.8 N - (2.0kg)(g)u = (2kg)(a). However i do not know how to solve for acceleration of A without the coefficient of friction
 
  • #4
It would help if you posted the figure that goes with this problem. Use the link button "Attach files" on the lower left.
 
  • #5
kuruman said:
It would help if you posted the figure that goes with this problem. Use the link button "Attach files" on the lower left.
attached in edit
 
  • #6
Thanks. Now consider this. How many forces act on the top block and what object(s) exert(s) them on it?
 
  • #7
kuruman said:
Thanks. Now consider this. How many forces act on the top block and what object(s) exert(s) them on it?
there is the gravitational force, the applied force, and the frictional force the applied force is applied by the bottom block
 
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  • #8
JMAMA said:
So my understanding is that from F = ma we can determine for block B that F = (6 kg)(1.80 ms2) = 10.80 N we can then say that block A is experiencing the same force yielding F - F(friction) = m a or 10.8 N - (2.0kg)(g)u = (2kg)(a). However i do not know how to solve for acceleration of A without the coefficient of friction
Welcome, @JMAMA !

You don’t need to calculate that force of friction in this case.
That pulling force is able to accelerate a mass of 6.66 kg at the given rate.
Therefore, it is simultaneously accelerating block B at that rate, and block A at a different rate.
 
  • #9
Draw the free body diagram for B. The method for solving the problem should then be apparent.
 
  • #10
Lnewqban said:
Welcome, @JMAMA !

You don’t need to calculate that force of friction in this case.
That pulling force is able to accelerate a mass of 6.66 kg at the given rate.
Therefore, it is simultaneously accelerating block B at that rate, and block A at a different rate.
what would the equation be for that the 0 friction is throwing me off. would it just be 6 m/s2 from 12N = 2kg a??
 
  • #11
Cutter Ketch said:
Draw the free body diagram for B. The method for solving the problem should then be apparent.
 

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  • #12
Well, sort of. A) What is that force to the left? Where does it come from. B) it’s not F=ma as you indicated for relevant equations, it’s
##\sum F## = ma so what should you do with that free body diagram?
 
  • #13
Outline:

1) Draw a FBD for block A

2) Write Newtons Second Law for block A

3) Draw FBD for Block B

4) Write Newtons Second Law for block B

5) Combine the two equations, solve for the acceleration of block ##A##

Where you might get stuck...the frictional force acting between blocks A and B. If it necessarily acts on block ##A## in a certain direction, in which way does it act on block ##B##?
 
  • #14
JMAMA said:
what would the equation be for that the 0 friction is throwing me off. would it just be 6 m/s2 from 12N = 2kg a??
Can you visualize block A accelerating at a lower rate than block B and falling behind?

If no friction existed between A and B, block A would not move respect to the ground, and our force of 12.0 N would induce a greater acceleration than 1.8 m/s2 to block B 9only one experiencing that force then).

If blocks A and B were solidly linked together, both blocks would move respect to the ground with identical acceleration, and our force of 12.0 N would induce a lower acceleration than 1.8 m/s2 to the A-B block system of mass 8.0 kg.
 

FAQ: Newton's law problem: Pushing 2 stacked blocks on a horizontal table

What is Newton's law problem involving pushing 2 stacked blocks on a horizontal table?

This problem typically involves analyzing the forces and accelerations acting on two blocks stacked on top of each other when a horizontal force is applied to one of the blocks. Using Newton's laws of motion, we aim to determine the resulting accelerations and the forces between the blocks.

How do you determine the acceleration of the blocks?

To determine the acceleration of the blocks, we need to consider the total mass of the system and the applied force. By applying Newton's second law (F = ma) to the entire system, we can find the acceleration by dividing the applied force by the total mass of both blocks.

What forces act on the blocks in this scenario?

The forces acting on the blocks include the applied force, the gravitational force (weight) of each block, the normal force from the table, the normal force between the blocks, and the frictional forces between the blocks and between the bottom block and the table.

How do you calculate the frictional force between the blocks?

The frictional force between the blocks can be calculated using the coefficient of friction (μ) and the normal force between the blocks. The normal force in this case is equal to the weight of the top block. The frictional force is given by F_friction = μ * N, where N is the normal force.

What happens if the applied force is too small to overcome friction?

If the applied force is too small to overcome the static friction between the blocks or between the bottom block and the table, the blocks will not move. In this case, the static frictional force will adjust to match the applied force, preventing any relative motion.

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