Force exerted by M1 on M2 problem

[JSmithDawg]

Homework Statement

The illustration is not drawn to scale. M2 = 4 kg and M1=20kg. They're both on a frictionless surface. A 36N constant force is applied to M2. What's the force exerted by m1 on m2?

Homework Equations

F=ma

The Attempt at a Solution

According to Newton's 3rd law, forces come in pairs. If there's 36N of force exerted on M2, the block should move and transfer the force onto M1. Since the surface is frictionless, I don't have to worry about any force being lost due to friction. Thus, 36N of force should be exerted onto M2 by M1. However, I know my logic is wrong, since the online quiz told me that the correct answer is 30N. How do I correctly approach this problem?

 

[haruspex]

According to Newton's 3rd law, forces come in pairs.

Yes, but as action and reaction. You can use that to determine the relationship between the forces the two masses exert on each other.
Consider the acceleration of the system. What force therefore acts on m1?

 

[Chestermiller]

Since M1 and M2 move together, if a force of 36 N is applied to them, what is their acceleration?

Chet

 

[Satvik Pandey]

Hi JSmithDawg.:)
Consider m1 and m2 as a system and find the acceleration of the system by using F=ma ,as Chestermiller said. Then draw free body diagrams of m1 and m2 separately. Can you find the magnitude of force acting on m1 due to m2? What is the relation between the acting on m1 by m2 and force acting on m2 by m1?

 

[HalfLostAndSearching]

Your logic is partially correct. According to Newton's 3rd law, the force exerted by M1 on M2 will be equal in magnitude but opposite in direction to the force exerted by M2 on M1. This means that the force exerted by M1 on M2 will also be 36N, but in the opposite direction.

To understand why the correct answer is 30N, we need to consider the forces acting on each block separately. Since M2 is being pushed with a constant force of 36N, it will have an acceleration of 9 m/s^2 (F=ma, 36N=4kg*9m/s^2). This means that M2 will have a net force of 36N acting on it in the direction of the applied force.

On the other hand, M1 is not being pushed or pulled by any external forces, so its net force is 0N. However, it is in contact with M2 and thus experiences a reaction force from M2 of 36N (equal in magnitude but opposite in direction). This means that M1 has a force of 36N acting on it in the direction opposite to the applied force on M2.

Now, since M1 has a mass of 20kg, its acceleration will be 1.8 m/s^2 (F=ma, 36N=20kg*1.8m/s^2). This means that M1 will be pushing on M2 with a force of 36N in the opposite direction of the applied force, but due to its lower acceleration, this force will have a magnitude of 1.8 m/s^2. Therefore, the force exerted by M1 on M2 is 30N (36N-1.8N).

In summary, the force exerted by M1 on M2 is 30N because M1 is pushing on M2 with a force of 36N, but due to its lower acceleration, the force has a magnitude of 1.8N less than the applied force on M2.