Magnitudes of Blocks on Surface

  • #1

Homework Statement



Three blocks are in contact with each other on a frictionless horizontal surface. A 402 N
horizontal force is applied to the block with mass of 2.9 kg as shown in the figure below.
The acceleration of gravity is 9.8 m/s2.

a. What is the net force on the block with mass 2.9 kg? Answer in units of N.
b. What is the resultant force on the block 2 with mass 5.4 kg? Answer in units of N
c. What is the resultant force on the block with mass 8.4 kg? Answer in units of N.
d. What is the magnitude of the force between the block with mass 5.4 kg and 8.4 kg? Answer in units of N.
e. What is the magnitude of the force between the block with mass 2.9 kg and 5.4 kg? Answer in units of N.

Homework Equations


F=ma


The Attempt at a Solution



I'm starting out by drawing the free body diagrams for each of the blocks. I'm using the +i to the right and +j to the top.

So far, I have:

2.9 kg block (1)
F(applied) = +Fi
F(gravity) = -m(1)gj
N(1) = +N(1)j

5.4 kg block (2)
F(gravity) = -m(2)gj
N(2) = +N(2)j

8.4 kg block (3)
F(gravity) = -m(3)gj
N(3) = +N(3)j

How do I put in the other forces for each block touching the one next to it?

Once I get the other normal forces, I can add the j and i for each block and I will have the net forces for each block.

How would I get d and e, the magnitudes of the force between the blocks?

Thanks!
 

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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Three blocks are in contact with each other on a frictionless horizontal surface. A 402 N
horizontal force is applied to the block with mass of 2.9 kg as shown in the figure below.
The acceleration of gravity is 9.8 m/s2.

How would I get d and e, the magnitudes of the force between the blocks?
Hi southernguy13! Welcome to PF! :smile:

Sorry, but I don't understand your solution at all. :redface:

(what does g have to do with it? :confused:)

Call the acceleration a, then write out good ol' Newton's second law for all three blocks together (that should give you the acceleration), and then write out Newton's second law for each block individually. :wink:
 
  • #3
It is confusing the way I'm trying to do it. This is just how I was taught in class, but obviously it isn't the easiest way.

F=ma
402 = (2.9+5.4+8.4)a
a=24.0719 m/s2

a. F=ma
F =2.9(24.0719)
=69.8084 N

b. F=ma
F= 5.4(24.0719)
=129.988

c. f=ma
F= 8.4 (24.0719)
=202.204

Those are all correct, I think.

So now how do I find the magnitude between the blocks?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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… So now how do I find the magnitude between the blocks?
Your F = ma for a. b. and c. is the net force on each block …

so it will be the total of the applied force F (first block only) and the force(s) between the blocks.

Try the right-hand block first. :smile:
 
  • #5
Hmm, I can't seem to get it right. The total applied force is 402 N. So for question d, I take 402 and subtract what? I tried 402 - 69.8084 but that isn't correct...
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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Hmm, I can't seem to get it right. The total applied force is 402 N. So for question d, I take 402 and subtract what? I tried 402 - 69.8084 but that isn't correct...
You're doing d, which is the force between the middle block and the right-hand block …

So use F = ma for the right-hand block: a is the acceleration you've already found, and F is the net force, which is just d. :wink:

Now try F = ma for the left-hand block. :smile:
 
  • #7
I got the last part, I added 129.988 and 202.204 and got 332.192. I just don't know how to get the 2nd to last one, between 5.4 kg and 8.4 kg. I just don't understand what I'm supposed to be finding. Sorry that I keep asking, I just really want to understand this.
 
  • #8
tiny-tim
Science Advisor
Homework Helper
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I got the last part, I added 129.988 and 202.204 and got 332.192. I just don't know how to get the 2nd to last one, between 5.4 kg and 8.4 kg. I just don't understand what I'm supposed to be finding. Sorry that I keep asking, I just really want to understand this.
I honestly have no idea what you're doing. :redface:

Can you write it out properly?

(and the 2nd to last one isn't between 5.4 kg and 8.4 kg)
 
  • #9
F=ma
402 = (2.9+5.4+8.4)a
a=24.0719 m/s2

a. F=ma
F =2.9(24.0719)
=69.8084 N

b. F=ma
F= 5.4(24.0719)
=129.988 N

c. f=ma
F= 8.4 (24.0719)
=202.204 N

d. I don't know how to find the answer.

e. I did 129.988 + 202.204 = 332.192 N. This is correct, but I don't know why.
 

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