Simple Mechanics Problem -- Block connected to a wall by a rope

In summary, the problem involves two blocks, A and B, with a negative acceleration observed in block B. Upon investigation, it is found that the force needed to move the system is slightly more than 54N. However, upon further analysis, it is discovered that equation 2 is incorrect and the correct equation is ##f_{k1}=\mu_k F_{N1}## with ##F_{N1}## being given by equation 1. This leads to the conclusion that the problem may be incorrect as a negative acceleration cannot be achieved with the given force.
  • #1
domingoleung
7
1
Homework Statement
Block A (5 kg) is connected to the wall with a cord of tension T, and initially at rest on top of block B (10 kg) on a rough floor. If there is an applied force 54 N acting on block B, block B could move to the right while block A is still at rest due to the connecting cord. Assume the coefficient of kinetic friction between blocks A and B and between block B and the ground is μk = 0.3.

(i) Draw free body diagrams of blocks A and B.
(ii) Find the tension T of the cord and the acceleration of block B.
Relevant Equations
F = ma
asm1.png
(So this is the system given)

The following is my analysis:
(i)
WhatsApp Image 2020-03-18 at 00.15.27.jpeg

(ii)
WhatsApp Image 2020-03-18 at 00.15.43.jpeg

Well, my problem is - I got a negative acceleration and its quite impossible to have block B moving to the left. So I am wondering if there are any mistakes I've made.
 
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  • #2
Can you write out the equations for all the forces on both blocks? Horizontal and vertical. In particular, can you check the calculation of ##T##. How did you get that?
 
  • #3
PeroK said:
Can you write out the equations for all the forces on both blocks? Horizontal and vertical. In particular, can you check the calculation of ##T##. How did you get that?

Okay, ##T## is correct.

I think I agree with you. You need a force of slightly more than ##54N## to get this system moving.

Increasing ##g## to ##10 m/s^2## makes things worse, of course.

You can check once more, but unless we've made the same mistake, I think the problem might be wrong. It's a bit suspicious, as ##54N## is approximately the force needed.
 
Last edited:
  • #4
Equation 2 is incorrect. ##f_{k1}=\mu_k F_{N1}## and ##F_{N1}## is given by equation 1.
 
  • #5
kuruman said:
##f_{k1}=\mu_k F_{N1}## and ##F_{N1}## is given by equation 1.

That's true, but considering horizontal forces on block ##A## also gives:

##f_{k1} = T\cos \theta##
 
  • #6
Ah yes. I forgot A is not accelerating. I agree that the 54 N force is not enough to provide an acceleration to the right. A negative acceleration could be compatible with the situation in which block B is already moving and then block A is dropped on it to slow it down, but that's not what the problem says.
 
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Related to Simple Mechanics Problem -- Block connected to a wall by a rope

1. What is the purpose of connecting the block to the wall with a rope in this simple mechanics problem?

The rope serves as a constraint or boundary condition for the block, allowing us to study its motion in a specific direction.

2. How does the tension in the rope affect the motion of the block?

The tension in the rope acts as a force on the block, causing it to accelerate or decelerate depending on the direction of the force and the mass of the block.

3. What factors determine the magnitude of the tension in the rope?

The magnitude of the tension in the rope depends on the weight of the block, the angle of the rope with respect to the wall, and any external forces acting on the block.

4. Can the block move in any direction if it is connected to the wall by a rope?

No, the block can only move in the direction of the rope due to the constraint it provides. However, the block can still rotate around the point where the rope is connected to the wall.

5. How can we calculate the acceleration of the block in this simple mechanics problem?

We can use Newton's second law, F=ma, to calculate the acceleration of the block. The net force acting on the block is equal to the mass of the block times its acceleration. We can also use trigonometry to break down the forces in the x and y directions.

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