A block of mass M1=2.9 kg rests on top of a second block of mass2=5.5 kg, and the second block sits on a surface that is so slippery that the friction can be assumed to be zero.
Question:
(a) if the coefficient of static friction between the blocks is 0.21, how much force can be applied to the top block without the blocks slipping apart?
(b) how much force can be applied to the bottom block for the same result?
The answer to part a above is incorrect. You must remember that the force of friction resisting the applied force on M1 also applies an equal and opposite force on M2. The applied force must overcome /both/ of these.
The sum of forces on the X axis for M2 is just this equal and opposite force (Force of friction). Setting this to ma will allow you to solve for an acceleration value.
You can use this acceleration value in solving for the applied force on M1. (T - Ffriction = ma).
Hope this helps.
(a) Top block:
F-friction=M1a
F=2.9a+(0.21*2.9*9.8)
Bottom block:
friction=M2a
a=(0.21*2.9*9.8)/5.5
F=(0.21*2.9*9.8*2.9)/5.5+(0.21*2.9*9.8)=9.12N
(b) Bottom block:
F-friction=M2a
F=5.5a+(0.21*2.9*9.8)
Top block: 0.21*2.9*9.8=2.9a
a=0.21*9.8
F=(0.21*5.5*9.8)+(0.21*2.9*9.8)=17.29N
To find the maximum force that can be applied without the blocks slipping apart, we need to consider the forces acting on the system.
Let's start with the top block (M1) on which the force will be applied.
(a) To find the maximum force that can be applied to the top block without the blocks slipping apart, we need to calculate the maximum static friction force between the blocks (Fs_max).
The formula for the maximum static friction force is:
Fs_max = μs * N
Where:
μs is the coefficient of static friction.
N is the normal force between the blocks.
In this case, the normal force (N) is equal to the weight of the top block (M1) which is given by:
N = M1 * g
Where:
g is the acceleration due to gravity which is approximately 9.8 m/s^2.
Now, substitute the values into the equation to get the maximum static friction force:
Fs_max = μs * N
= μs * M1 * g
Substitute the given values into the equation:
Fs_max = 0.21 * 2.9 kg * 9.8 m/s^2
Calculate the result:
Fs_max = 6.0834 N
Therefore, the maximum force that can be applied to the top block without the blocks slipping apart is approximately 6.0834 N.
(b) To find the maximum force that can be applied to the bottom block for the same result, we need to consider the forces acting on the bottom block (M2).
Since the surface between the bottom block and the surface is assumed to be frictionless, the only force acting on the bottom block is its weight.
The weight of the bottom block (M2) is given by:
W = M2 * g
Substitute the given values into the equation:
W = 5.5 kg * 9.8 m/s^2
Calculate the result:
W = 53.9 N
Therefore, the maximum force that can be applied to the bottom block without the blocks slipping apart is equal to the weight of the bottom block which is approximately 53.9 N.
(a) 0.21*M1*g = 5.968 N
(b) The maximum friction force that M2 can apply to M1 without M1 slipping is also 5.968 N. That will cause M1 AND M2 to accelerate at rate
a = 5.968 N/M1 = 2.058 m/s^2
The force that would have to be applied to M2 to make this happen is
(M1 + M2)*a = 8.4 * 2.058 = 17.29 N