What force is Block A exerting on Block B?

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The discussion revolves around calculating the force exerted by Block A on Block B in a physics problem involving two blocks with different masses and coefficients of friction. Participants emphasize the importance of identifying all forces acting on both blocks, including applied forces and friction, to determine the system's acceleration. A free body diagram is recommended to visualize the forces on each block, which aids in applying Newton's laws to find the unknown force. An example calculation is provided, using specific values for mass, friction coefficients, and applied force, leading to a calculated force of approximately 166.64 N exerted by Block A on Block B. The conversation highlights the necessity of ensuring both blocks maintain contact and move with the same acceleration for accurate calculations.
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Homework Statement



This is a question that the teacher gives every week that we do in class. He gives the parameters without the actual numbers a week before so we can prepare.

The teacher will give:

- the masses for both blocks (different)
- the coefficient of friction for both blocks (different)
- the applied force on Block A

Question (what we have to figure out): What force is Block A exerting on Block B?

http://img580.imageshack.us/img580/7968/85935850.png"

Homework Equations



Symbols weren't working for me but I'm assuming these equations are relevant:

coefficient of friction = Net Force / Normal Force

and

Fnet = mass * acceleration

The Attempt at a Solution



I'm not sure how to go about at this. I would have to specify some numbers before I try. I would have to find all the forces acting on both objects first (one force would be the mass, one force would be the friction, and one would be the applied force for Block B). After that, I have no idea what to do.
 
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rarouch said:
Question (what we have to figure out): What force is Block A exerting on Block B?


Homework Equations



Symbols weren't working for me but I'm assuming these equations are relevant:

coefficient of friction = Net Force / Normal Force

and

Fnet = mass * acceleration

The Attempt at a Solution



I'm not sure how to go about at this. I would have to specify some numbers before I try. I would have to find all the forces acting on both objects first (one force would be the mass, one force would be the friction, and one would be the applied force for Block B). After that, I have no idea what to do.
When you look at both objects sliding together as a system, you have, in the x direction, the applied forces and the friction forces. Mass is not a force. Knowing these forces and the mass, you can calculate the acceleration of the blocks as they move together with the same acceleration.
To calculate the force between the blocks, you have to draw a free body diagram of one of the blocks and identify all forces acting on it, then apply Newton's law again to that block to solve for the unknown force. You should look at the other block also in the same manner, as a check on your work.
 
Alright, so I tried specifying numbers on my own. Here's my go at it:

Block A mass = 8 kg
Block B mass = 20 kg
Coefficient of kinetic friction for Block A = 0.20
Coefficient of kinetic friction for Block B = 0.40
Applied Force = 200 N

Find Normal Force
Block A
F = ma
F = 8kg x 9.8
F = 78.4 N
F = (78.4N)(0.20) - taking coefficient of friction into consideration
F = 15.68 N
--------------------
Block B
F = ma
F = 20kg x 9.8
F = 196 N
F = (196N)(0.40)
F = 78.4 N
-------------------
Find Acceleration
Total normal force = 78.4 N + 15.68 N = 94.1 N
Convert to kg = 94.1 N/9.8 = 9.60 kg
------------------
Acceleration
a = F/m
a = 200N/9.6kg
a = 20.83m/s^2 [right, assuming applied force is to the right]
------------------
Force of Block A on Block B
F = ma
F = (8kg)(20.83m/s^2)
F = 166.64 N [E]

I know I completely messed up one way or another. Would this be the right way, if not, can someone please correct me?
 
I probably didn't do this right.
 
try to draw a free body diagram of both the masses seperately. assume the accln. to be the same for both the masses, because otherwise they would break contact (i.e. if B has more acln. than that of A, also think why it is not possible that B has less accln. than that of A) and hence there will be no force btwn. them.

the forces on A:

A1. applied force F
A2. Friction opposing the motion.
A3. force exerted by B on A, also opposing motion.

add the forces vectorially and equate the result with mass X accln. of A.

the forces on B:
B1. Friction opposing the motion.
B2. force exerted by A on B, this time favouring the motion.
add the forces vectorially and equate the result with mass X accln. of B.

now from Newton's 3rd law put A3 = B2

now u have 3 eqn.s with three unknowns, assuming the accln. of the masses to be same. so u can solve it.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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