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My teacher said there isn't a special kind of formula, you just derive it from f=ma.
This is the diagram:
This is the diagram:
I assume you want the contact force between those two masses. Your teacher is right. Just apply Newton's 2nd law.
Hint: First find the acceleration of the system.
Good. Now apply Newton's 2nd law to M1 alone.Yes. I found the acceleration.
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.
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 ?
When applying Newton's 2nd law to m1, use the mass of m1.Why would you use the mass of m1 and not m2 ?
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 the net force: ƩF = m2*a.If I do 1.32 m/s^2 *3.46 m/s^2 I don't get the same answer...
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 ?
Sure. First identify the forces acting on each block. What forces act on m2? On m1?
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.
Only one force acts on m1. In what direction does it act?<-m1
<-m2->
or
<-m1->
<-m2->
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
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 ?
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 ?
<---(Fc)---[m2]---(F)--->