Force on Two Blocks: Find F, Acceleration

In summary, the assembly of blocks with a mass of 4.0 kg on top of a block with a mass of M = 5.5 kg requires a horizontal force of at least 18 N to cause the top block to slip on the bottom one while the bottom one is held fixed. The maximum horizontal force F that can be applied to the lower block for the blocks to move together is 18 N. The resulting acceleration of the blocks is 4.5 m/s^2. The frictional force acts on the top block in the direction of motion and on the bottom block in the opposite direction. The maximum frictional force is 18 N, and when applied, it results in the same acceleration for both blocks.
  • #1
mbrmbrg
496
2
A block of mass 4.0 kg is put on top of a block of mass M = 5.5 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 18 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.

(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.

(b) Find the magnitude of the resulting acceleration of the blocks.

Once the blocks are on the frictionless table, I run into trouble.
Namely, I don't know where to place friction in my free-body-diagrams for the two blocks. Friction definitely acts on the upper block, and I think that it points in the same direction as the force applied to the bottom block. It will also probably be maximum frictional force, which was given as 18N.
But does that friction act on the bottom block? I want to say not, but it's a contact force, and the bottom block is after all in contact with the top block!
 
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  • #2
Something odd this way comes:
I worked with the assumption that in the x direction, [tex]\Sigma F_B=F=m_Ba[/tex] (where F=applied force) and [tex]\Sigma F_A=f=m_Aa[/tex] (where f=maximum static fictional force between the two blocks).
I subtracted the second equation from the first and got [tex]\frac{F}{m_B}-\frac{f}{m_A}=0[/tex] to get F=24.75N.
That is the wrong answer, but I didn't know that yet, so I plugged F=24.75 into my first equation to get a=4.5m/s^2. Lo and behold! The a is right even though the F is wrong!
Any ideas?
 
  • #3
Going on the off chance that [tex]\Sigma F_B=F+f[/tex], I subtracted 18 from 24.75. That's still the wrong answer...
 
  • #4
mbrmbrg said:
A block of mass 4.0 kg is put on top of a block of mass M = 5.5 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 18 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.

(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.

(b) Find the magnitude of the resulting acceleration of the blocks.

Once the blocks are on the frictionless table, I run into trouble.
Namely, I don't know where to place friction in my free-body-diagrams for the two blocks. Friction definitely acts on the upper block, and I think that it points in the same direction as the force applied to the bottom block. It will also probably be maximum frictional force, which was given as 18N.
But does that friction act on the bottom block? I want to say not, but it's a contact force, and the bottom block is after all in contact with the top block!
You are not far off. 18N is the maximum force that can be applied before moving the top block while holding the bottom block, so that is the maximum frictional force. When maximum F is applied to the lower block, the friction force between the two blocks will be 18N. Friction acts one way on the top block (direction of motion) and the opposite way on the bottom block. It will oppose the F applied to the lower block and it will result in accelerating the upper block. From the upper block you find the acceleration of both blocks. What force is strong enough to overcome the 18N it has to fight, and still accelerate the lower block the same as the upper block?

An easier way to look at it is that since both blocks are moving with the same acceleration they are acting as a single mass responding to the applied F. What force will give the total mass the acceleration of the upper block? Both points of view give the same result.
 
Last edited:
  • #5
Neat-o!
Thanks OlderDan.
 

Related to Force on Two Blocks: Find F, Acceleration

1. What is the formula for calculating force on two blocks?

The formula for calculating the force on two blocks is F = m1 x a1 + m2 x a2, where F is the total force, m1 and m2 are the masses of the blocks, and a1 and a2 are the respective accelerations of the blocks.

2. How do I find the acceleration of the blocks?

To find the acceleration of the blocks, you first need to determine the net force acting on the system. This can be done by adding up all the individual forces acting on the blocks. Then, use Newton's second law of motion (F = m x a) to calculate the acceleration.

3. What if the blocks have different masses?

If the blocks have different masses, you will need to use the formula F = m1 x a1 + m2 x a2 to calculate the total force. However, the acceleration of each block will be different, depending on its mass and the net force acting on it.

4. Can the force on the blocks be negative?

Yes, the force on the blocks can be negative. This indicates that the force is acting in the opposite direction of the positive direction chosen for the calculation. It is important to pay attention to the direction of the forces and accelerations in order to get an accurate result.

5. How can I use this information in real-life scenarios?

The concept of force on two blocks can be applied to various real-life scenarios, such as determining the force needed to push or pull an object, calculating the acceleration of a car or train, and understanding the forces involved in a collision. It is a fundamental concept in physics and is used in many practical applications, from engineering to sports.

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