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This is the diagram:

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- Thread starter Taz.
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- #1

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This is the diagram:

- #2

Doc Al

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Hint: First find the acceleration of the system.

- #3

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Hint: First find the acceleration of the system.

Yes. I found the acceleration.

- #4

Doc Al

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Good. Now apply Newton's 2nd law to M1 alone.Yes. I found the acceleration.

- #5

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So-> F= m1 * acceleration that I found for the whole system ?

Ok, I found the answer. But scientifically, I don't understand how.

My teacher said that the contact force for both blocks is the same. How is that possible ? So if a bee flying at 10 km/h hits a car moving at 30 km/h the forces between both bodies is the same. How is that possible ?

- #6

Doc Al

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Right.To find the the answer in N ?

So-> F= m1 * acceleration that I found for the whole system ?

That's Newton's 3rd law. Whenever two objects interact, they exert equal and opposite forces on each other.Ok, I found the answer. But scientifically, I don't understand how.

My teacher said that the contact force for both blocks is the same. How is that possible ?

Definitely.So if a bee flying at 10 km/h hits a car moving at 30 km/h the forces between both bodies is the same.

- #7

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Ok but why did we use m1 rather then m2 when we applied newtons second law on m1 ?

- #8

Doc Al

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If you're applying Newton's 2nd law to m1, why in the world would you use m2?Ok but why did we use m1 rather then m2 when we applied newtons second law on m1 ?

- #9

SammyS

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Draw two Free Body Diagrams, one for each block

- #10

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We found acceleration for the whole system.

Then the questions asked:

Find the magnitude of the contact force between the two blocks.

Why would you use the mass of m1 and not m2 ?

- #11

Doc Al

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When applying Newton's 2nd law to m1, use the mass of m1.Why would you use the mass of m1 and not m2 ?

You can also apply Newton's 2nd law to m2, in which case you'll use the mass of m2.

You'll get the same answer for the contact force either way. (Analyzing m1 is a bit easier, since only one force acts.)

- #12

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My acceleration is: 1.32 m/s^2

And then I just multiplied 1.32 m/s^2 by 1.39 m/s^2.

If I do 1.32 m/s^2 *3.46 m/s^2 I don't get the same answer...

- #13

Doc Al

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Right.My acceleration is: 1.32 m/s^2

That's correct.And then I just multiplied 1.32 m/s^2 by 1.39 m/s^2.

The contact force is not the only force acting on m2. To apply Newton's 2nd law to m2, you must use theIf I do 1.32 m/s^2 *3.46 m/s^2 I don't get the same answer...

- #14

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I get it. Thanks.

Can you help me to draw a FBD for this problem ?

Can you help me to draw a FBD for this problem ?

- #15

Doc Al

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Sure. First identify the forces acting on each block. What forces act on m2? On m1?Can you help me to draw a FBD for this problem ?

- #16

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Sure. First identify the forces acting on each block. What forces act on m2? On m1?

Not sure but

m1-> force coming from m2 and force going forward.

m2->force from F (6.40N) and force of contact.

- #17

Doc Al

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What do you mean by 'force going forward'? The only thing in contact with m1 is m2.m1-> force coming from m2 and force going forward.

Right. They act in different directions, of course.m2->force from F (6.40N) and force of contact.

- #18

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<-m1

<-m2->

or

<-m1->

<-m2->

<-m2->

or

<-m1->

<-m2->

- #19

Doc Al

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Only one force acts on m1. In what direction does it act?<-m1

<-m2->

or

<-m1->

<-m2->

- #20

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Well there is the force of contact... Does this count ?

- #21

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Thanks !

but just a last question-> for m2 is it 6.20 + F = 2.82 *1.18 OR -6.20 + F = 2.82 *1.18

- #22

Doc Al

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I don't quite understand what you are doing here. You want:but just a last question-> for m2 is it 6.20 + F = 2.82 *1.18 OR -6.20 + F = 2.82 *1.18

ƩF = m2*a

The two forces are the applied force "F" which acts to the right and the contact force "Fc" which acts to the left. If you choose a sign convention so that the right is positive, then you have:

+F -Fc = m2*a

6.2 - Fc = 3.46*1.32

And so on.

- #23

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How would the FBD look like for m2 ?

- #24

Doc Al

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The applied force F acting to the right; the contact force from m1 acting to the left.How would the FBD look like for m2 ?

- #25

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But where are the arrow pointing ?

<-m2->

Like this ?

<-m2->

Like this ?

- #26

Doc Al

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That doesn't specify which arrow goes with which force. I'd draw it like this:But where are the arrow pointing ?

<-m2->

Like this ?

Code:

`<---(Fc)---[m2]---(F)--->`

- #27

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ok sounds good. Thanks for your help. Very good teaching skills.

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