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bp789
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1. Homework Statement
There is an attachment of the picture of the problem.
The figure shows three crates being pushed over a concrete floor by a horizontal force F of magnitude 647 N. The masses of the crates are m1 = 28 kg, m2 = 12 kg, and m3 = 22 kg. The coefficient of kinetic friction between the floor and each of the crates is 0.7. What is the magnitude F32 of the force on crate 3 from crate 2?
2. Homework Equations
F=ma
Ffric=μN
Fg=mg
3. The Attempt at a Solution
Okay, so the first thing I did was find the gravitational forces of all three crates, which would also mean finding the normal force of each of the three crates since the gravitational and normal force cancel each other out, and there aren't any other forces in the vertical direction.
Crate 1: Fg=(28 kg)(9.8 m/s2)=274.4 N
Crate 2: Fg=(12 kg)(9.8 m/s2)=117.6 N
Crate 3: Fg=(22 kg)(9.8 m/s2)=215.6 N
Next, I found all of the forces of friction of the crates
Crate 1: Ffric=(0.7)(274.4 N)=192.08 N
Crate 2: Ffric=(0.7)(117.6 N)=82.32 N
Crate 3: Ffric=(0.7)(215.6 N)=150.92 N
Next thing I did was to find the total net force by subtracting the forces of friction from the horizontal force given (647 N), and then use that number to find the acceleration. We know that a 647 N is the horizontal force given and is pushing the system so that force is also included in the net force.
Fnet=(647-192.08-82.32-150.92)=221.68
Fnet=ma
221.68=(28+12+22)a
221.68=62a
a=3.575 m/s2
Now according to the tutorial of the question in my online textbook, all of this above information is correct and we have everything we need to find the answer.
I assumed that the next thing to do was to multiply the acceleration we found by the mass of crate 3 because we are looking for the magnitude of the force on crate 3 by crate 2. I did:
F=(22 kg)(3.575 m/s2)=78.66 N
However, when I checked my answer, it said it was wrong. The program doesn't give me the correct answer. I have to figure it out for myself and then it will tell me whether it is correct or not, but I'm not sure what I did wrong here?...
There is an attachment of the picture of the problem.
The figure shows three crates being pushed over a concrete floor by a horizontal force F of magnitude 647 N. The masses of the crates are m1 = 28 kg, m2 = 12 kg, and m3 = 22 kg. The coefficient of kinetic friction between the floor and each of the crates is 0.7. What is the magnitude F32 of the force on crate 3 from crate 2?
2. Homework Equations
F=ma
Ffric=μN
Fg=mg
3. The Attempt at a Solution
Okay, so the first thing I did was find the gravitational forces of all three crates, which would also mean finding the normal force of each of the three crates since the gravitational and normal force cancel each other out, and there aren't any other forces in the vertical direction.
Crate 1: Fg=(28 kg)(9.8 m/s2)=274.4 N
Crate 2: Fg=(12 kg)(9.8 m/s2)=117.6 N
Crate 3: Fg=(22 kg)(9.8 m/s2)=215.6 N
Next, I found all of the forces of friction of the crates
Crate 1: Ffric=(0.7)(274.4 N)=192.08 N
Crate 2: Ffric=(0.7)(117.6 N)=82.32 N
Crate 3: Ffric=(0.7)(215.6 N)=150.92 N
Next thing I did was to find the total net force by subtracting the forces of friction from the horizontal force given (647 N), and then use that number to find the acceleration. We know that a 647 N is the horizontal force given and is pushing the system so that force is also included in the net force.
Fnet=(647-192.08-82.32-150.92)=221.68
Fnet=ma
221.68=(28+12+22)a
221.68=62a
a=3.575 m/s2
Now according to the tutorial of the question in my online textbook, all of this above information is correct and we have everything we need to find the answer.
I assumed that the next thing to do was to multiply the acceleration we found by the mass of crate 3 because we are looking for the magnitude of the force on crate 3 by crate 2. I did:
F=(22 kg)(3.575 m/s2)=78.66 N
However, when I checked my answer, it said it was wrong. The program doesn't give me the correct answer. I have to figure it out for myself and then it will tell me whether it is correct or not, but I'm not sure what I did wrong here?...
