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Fnet = ma problem

  • Thread starter A_lilah
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  • #1
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Homework Statement


Blocks A and B are sitting side by side (touching) on a horizontal surface. A constant horizontal force Fa is applied to block A, which pushes against block B with a 32.0 N force directed horizontally to the right. The same force Fa is applied to block B; now block A pushes on block B with a 16.0 N force directed horizontally to the left. The blocks have a combined mass of 30.0 kg.

(a) What is the magnitude of the acceleration of the blocks in figure (a)?
(b) What is the magnitude of the force Fa?

Homework Equations



Fnet = ma

The Attempt at a Solution




Part 1: Fa pushes on block A:
Block A has Fa pushing to the right, and block B pushing to the left so

Fnet = massblockA *acceleration = Fa - 32N


Block B pushes back on block A (only horizontal force on it)

Fnet = massblockB *acceleraton = 32N

mass blockA + massblockB = 30kg

Using these three equations I solved for Fa and got that Fa = 30kg*acceleration...
I thought that I could use the next set of data (Fa pushing on block B) to come up with another equation, then I could solve for Fa and acceleration, but the equations are exactly the same, so pretty useless.
Any help would be great, thanks!
(ps~ I sometimes have trouble with FBDs, so if something is wrong with how I laid out the forces, it would be great to know!)
 

Answers and Replies

  • #2
Doc Al
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I thought that I could use the next set of data (Fa pushing on block B) to come up with another equation, then I could solve for Fa and acceleration, but the equations are exactly the same, so pretty useless.
They aren't exactly the same. Write an equation for the force on block A using the second set of data.
 
  • #3
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Ok, Now I'm really not so sure:
For Fa pusing on block A:

massA + massB = 30kg
massblockA *acceleration = Fa - 32N
massblockB *acceleraton = 32N

so,

(Fa - 32)/ acceleration = massA
32/acceleration = mass B

(Fa - 32) + 32 = 30 * acceleration
So Fa = 30 * acceleration

For Fa pusing on B:

massA = 16N / acceleration
masB = (Fa -16) / acceleration
16N + (Fa - 16) = 30 * acceleration
or, Fa = 30 * acceleration

.... am I having a math problem then?
 
  • #4
Doc Al
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44,880
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Ok, Now I'm really not so sure:
For Fa pusing on block A:

massA + massB = 30kg
massblockA *acceleration = Fa - 32N
massblockB *acceleraton = 32N

so,

(Fa - 32)/ acceleration = massA
32/acceleration = mass B

(Fa - 32) + 32 = 30 * acceleration
So Fa = 30 * acceleration

For Fa pusing on B:

massA = 16N / acceleration
masB = (Fa -16) / acceleration
16N + (Fa - 16) = 30 * acceleration
or, Fa = 30 * acceleration

.... am I having a math problem then?
Combine the two equations I highlighted.
 
  • #5
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Oh.... how did you know to use those equations? (I have a final coming up, and any tips would be great)
Thanks for your help!
 
  • #6
Doc Al
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Oh.... how did you know to use those equations?
To me, those two equations are the most direct mathematical statements of the given data:

Blocks A and B are sitting side by side (touching) on a horizontal surface. A constant horizontal force Fa is applied to block A, which pushes against block B with a 32.0 N force directed horizontally to the right.
That tells me that: 32.0 = massB*a

The same force Fa is applied to block B; now block A pushes on block B with a 16.0 N force directed horizontally to the left.
From Newton's 3rd law, I immediately know the force on block A is 16.0 N.
That tells me that: 16.0 = massA*a

Add those equations to get:
32 + 16 = (massA + massB)*a

Then apply:
The blocks have a combined mass of 30.0 kg.
To get the acceleration.

The best thing to do is to write clear FBD diagrams of both blocks for both sets of data. Write all the equations and look for any easy ways to combine them.
 
  • #7
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Thanks for your help :)
 
  • #8
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Combine the two equations I highlighted.
Sorry if I bring it up, but I have the same problem, and I'm trying to figure out how to solve it. I read all the post above, and one thing I don't understand is, the acceleration in those 2 figures should be different, they can't be the same, and if they're not the same, how can we set up the equations ?
 
  • #9
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unloved .... did you draw FBD ?

EDIT:first assume that the acceleration is different in two cases, when you set up equations , you will see that they turn out to be same
 
Last edited:
  • #10
Doc Al
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Sorry if I bring it up, but I have the same problem, and I'm trying to figure out how to solve it. I read all the post above, and one thing I don't understand is, the acceleration in those 2 figures should be different, they can't be the same, and if they're not the same, how can we set up the equations ?
Why do you think the accelerations of the two blocks are different? They are in constant contact.
 

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