Force on two touching blocks

[Heather_]

Homework Statement

For two touching blocks, a constant horizontal force Fa is applied to block A to the right, which pushes against block B with a 26.0 N force directed horizontally to the right. In figure (b), the same force Fa is applied to block B to the right (the blocks switched places); now block A pushes on block B with a 13.0 N force directed horizontally to the left. The blocks have a combined mass of 6.00 kg (A is smaller than B).

I need to know the system's total acceleration and the force being applied.

Homework Equations

Newton's laws; friction is being ignored

The Attempt at a Solution

is it safe to assume that the mass of A is half the mass of B? (a pushes on b with half the force that b forces on a). That would make the acceleration 52, which isn't correct. Have i missed something?

 

[PhanthomJay]

Hi Heather, welcome to Physics Forums!

is it safe to assume that the mass of A is half the mass of B? (a pushes on b with half the force that b forces on a).

Yes, this is a very good observation. Now since A is half of B, and together their mass is 6 kg, then A = ? kg and B = ? kg.

That would make the acceleration 52, which isn't correct. Have i missed something?

Please show how you arrived at this number. Draw free body diagrams of each block for each case, and a free body diagram of the system of 2 blocks.

 

[Heather_]

Well obviously that would make the mass of A 2 kg and the mass of B 4. My thinking was that the block that was being pushed by the other one would need to exert an opposite force, so if A is pushing on B with a 26 N force and weighs 2 kg, the acceleration would be 13, which doesn't work (if A pushes on B with 26N force, though, why isn't that the force being applied to the system). As for the 52...I understand now why that was wrong.

 

[PhanthomJay]

Well obviously that would make the mass of A 2 kg and the mass of B 4. My thinking was that the block that was being pushed by the other one would need to exert an opposite force, so if A is pushing on B with a 26 N force and weighs 2 kg, the acceleration would be 13, which doesn't work[

If , drawing a free body diagram of B, A pushes on B with 26 N, and B has a mass of 4 kg, then per Newton 2, a = ?

(if A pushes on B with 26N force, though, why isn't that the force being applied to the system).

The force applied to the system is not the same as the force between blocks. The force applied to the system (the applied force acting on A in your first case) must move BOTH blocks.

 

[asteriatic]

As a scientist, it is important to clarify any assumptions made in a problem before proceeding with a solution. In this case, it is not safe to assume that the mass of block A is half the mass of block B. The problem states that the blocks have a combined mass of 6.00 kg, but does not specify the individual masses of each block. Therefore, it is necessary to use the given information and equations to solve for the individual masses of blocks A and B before calculating the total acceleration and force being applied.

To solve for the individual masses, we can use Newton's second law, F=ma, where F is the net force, m is the mass, and a is the acceleration. In the first scenario, where block A is pushing against block B, the net force on the system is Fa - 26.0 N, and the acceleration is unknown. Therefore, we can set up the equation Fa - 26.0 N = (m_A + m_B)a, where m_A and m_B are the masses of blocks A and B, respectively. Similarly, in the second scenario, where block B is pushing against block A, the net force is Fa - 13.0 N, and the acceleration is also unknown. Setting up the same equation, Fa - 13.0 N = (m_A + m_B)a.

Solving these two equations simultaneously, we can find that m_A = 2.00 kg and m_B = 4.00 kg. Now, we can use these values to calculate the total acceleration of the system using the first equation, Fa - 26.0 N = (2.00 kg + 4.00 kg)a. This gives us an acceleration of a = 4.33 m/s^2. Finally, we can use this acceleration value and the equation F=ma to solve for the force being applied, which is Fa = (2.00 kg + 4.00 kg)(4.33 m/s^2) = 26.00 N. Therefore, the system's total acceleration is 4.33 m/s^2 and the force being applied is 26.00 N.

In conclusion, it is important to carefully analyze the given information and use appropriate equations to solve for the total acceleration and force being applied in a system. Making assumptions without proper justification can lead to incorrect solutions.