Frictional Force: Find Accel. of 3kg & 2kg Blocks

In summary, the conversation discusses the problem of finding the acceleration of two blocks - one with a mass of 3.00kg and the other with a mass of 2.00kg - on a rough horizontal surface when a force of 29.0 N is applied to the 3.00kg block. The coefficient of friction between the blocks is given as \mu_s=0.35 and the kinetic friction between the 3.00kg block and the table is \mu_k=0.47. The conversation explores different approaches to finding the acceleration, with the correct conclusion being that both blocks will have the same acceleration of 1.194m/s^2.
  • #1
ubiquinone
43
0
Question: A force of [tex]29.0 N[/tex] is applied to a [tex]3.00kg[/tex] block which is initially at rest on a rough horizontal surface. A [tex]2.00kg[/tex] block is initially at rest on the [tex]3.00kg[/tex] block. The coefficient of friction between the blocks is [tex]\mu_s=0.35[/tex] and the kinetic friction between the [tex]3.00kg[/tex] block and the table is [tex]\mu_k=0.47[/tex].
a) Find the acceleration of the [tex]3kg[/tex] block.
b) Find the acceleration of the [tex]2kg[/tex] block.
 
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  • #2
I think, One thing you could do is that, find the force of friction and then find then thr net force, after that since you have the net force and the mass, you can find acceleration.
 
  • #3
Would it be something like this:
Let the applied force of 29N be F.
a) For the 3.0kg block, the frictional force between it and the table = [tex]\mu_k(3.0kg+2.0kg)g=0.470(5g)N[/tex]
The frictional force between the blocks also act upon the 3.0kg block = [tex]0.38(2g)N[/tex]
Writing, [tex]F_{net}[/tex] for the 3.0kg block, we have [tex]29N-0.38(2g)-0.470(5g)=3a[/tex]
Solving for [tex]a[/tex], we have [tex]a=-0.49m/s^2[/tex]
b) For the 2.0 kg block, the only force affecting its motion is the frictional force between it and the 3.0 kg block. That frictional force = [tex]0.38(2g)N[/tex]. So [tex]F_{net}=0.38(2g)N=2a[/tex], solving for [tex]a[/tex], we have [tex]a=3.7m/s^2[/tex]

Please, correct me if I am wrong, it would help me greatly. Thanks!
 
  • #4
ubiquinone said:
Would it be something like this:
Let the applied force of 29N be F.
a) For the 3.0kg block, the frictional force between it and the table = [tex]\mu_k(3.0kg+2.0kg)g=0.470(5g)N[/tex]
The frictional force between the blocks also act upon the 3.0kg block = [tex]0.38(2g)N[/tex]
Writing, [tex]F_{net}[/tex] for the 3.0kg block, we have [tex]29N-0.38(2g)-0.470(5g)=3a[/tex]
Solving for [tex]a[/tex], we have [tex]a=-0.49m/s^2[/tex]
b) For the 2.0 kg block, the only force affecting its motion is the frictional force between it and the 3.0 kg block. That frictional force = [tex]0.38(2g)N[/tex]. So [tex]F_{net}=0.38(2g)N=2a[/tex], solving for [tex]a[/tex], we have [tex]a=3.7m/s^2[/tex]

Please, correct me if I am wrong, it would help me greatly. Thanks!
There are some errors here. Does it make sense to you that the 3kg block has negative acceleration? Does it make sense that the 2kg has a positive acceleration while the 3kg block has a negative aceleration?

I'm not sure if the .35 you state in the problem or the 0.38 you use in the computation is the correct number. Be sure to check this. Either way, you cannot assume that the actual force of friction between the blocks is the maximum possible force. If the blocks do not slip, the force can be much less than the force you calculated.
 
  • #5
Alright thanks a lot OlderDan for replying.

I rethought of the problem and may someone please tell me if my logic now is correct.

The static friction between the blocks is 0.35, the kinetic friction between the 3.00kg block and the surface is 0.47, we can find kinetic friction first.

[tex]F_{kf}=\mu_k(m_1+m_2)g=0.47(5.00kg)(9.8m/s^2)=23.03N[/tex]
So the horizontal applied force is [tex]29N-23.03N=5.97N[/tex]

Now we calculate the static friction between the blocks, if it is greater than [tex]5.97N[/tex], then the [tex]2.00kg[/tex] block on top does not move.

[tex]F_{sf}=\mu_s(m_2)g=0.35(2.00kg)(9.8m/s^2)=6.86N[/tex]. Because [tex]6.86N>5.97N[/tex], the [tex]2.00kg[/tex] block does not move.

So we can treat the two blocks as one massive block with mass of [tex]5.00kg[/tex]. Then [tex]5.97N=(5.00kg)a[/tex] and [tex]a=1.194m/s^2[/tex]

Part b) wouldn't the [tex]2.00kg[/tex] and the [tex]3.00kg[/tex] mass then have the same acceleration of [tex]1.194m/s^2[/tex]?
 
  • #6
ubiquinone said:
Alright thanks a lot OlderDan for replying.

I rethought of the problem and may someone please tell me if my logic now is correct.

The static friction between the blocks is 0.35, the kinetic friction between the 3.00kg block and the surface is 0.47, we can find kinetic friction first.

[tex]F_{kf}=\mu_k(m_1+m_2)g=0.47(5.00kg)(9.8m/s^2)=23.03N[/tex]
So the horizontal applied force is [tex]29N-23.03N=5.97N[/tex]

Now we calculate the static friction between the blocks, if it is greater than [tex]5.97N[/tex], then the [tex]2.00kg[/tex] block on top does not move.

[tex]F_{sf}=\mu_s(m_2)g=0.35(2.00kg)(9.8m/s^2)=6.86N[/tex]. Because [tex]6.86N>5.97N[/tex], the [tex]2.00kg[/tex] block does not move.

So we can treat the two blocks as one massive block with mass of [tex]5.00kg[/tex]. Then [tex]5.97N=(5.00kg)a[/tex] and [tex]a=1.194m/s^2[/tex]

Part b) wouldn't the [tex]2.00kg[/tex] and the [tex]3.00kg[/tex] mass then have the same acceleration of [tex]1.194m/s^2[/tex]?
This is the correct conclusion, but the comparison of forces that led you to it is not quite right. The 6.86N is the maximum possible force on the 2kg block, but the 5.97N force is the net force on the combined 3kg+2kg system assuming the 2kg block does not slide. What you should be comparing to verify that the 2kg block does not slide is the maximum possible force to the force required to give the 2kg block the acceleration you computed. The net force on the system could have been greater than 6.86N and the 2kg block still not slide.

I know what you meant, but it is not correct to say the 2kg block does not move. What you mean is that it does not move relative to the 3kg block or does not slide. To an observer watching the blocks react to the applied force, the 2kg block will certainly move.
 

1. What is frictional force?

Frictional force is the force that acts against the motion of an object, caused by the contact between two surfaces.

2. How is frictional force calculated?

Frictional force can be calculated by multiplying the coefficient of friction between two surfaces by the normal force acting on the object.

3. What is the relationship between mass and frictional force?

There is no direct relationship between mass and frictional force. However, the weight of an object, which is affected by its mass, can contribute to the normal force and thus affect the frictional force.

4. How do you find the acceleration of two blocks with different masses connected by a rope?

To find the acceleration, we can use the equation F = ma, where F is the net force acting on the system, m is the total mass of the blocks, and a is the acceleration. The net force can be calculated by subtracting the frictional force from the tension force of the rope. Once we have the net force, we can solve for the acceleration.

5. Can frictional force be reduced or eliminated?

Frictional force can be reduced by using lubricants or smoother surfaces between the two objects. In some cases, it can also be eliminated by reducing the normal force acting on the object. However, frictional force cannot be completely eliminated as it is a fundamental force in nature.

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