Recent content by Taz.

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

So I have w1 and w2 and I have m1 and m2. Whats missing is T and A. Can I just solve for A or T ?

But where are the arrow pointing ? <-m2-> Like this ?

How would the FBD look like for m2 ?

I GOT IT ! 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

m2: force of tension going up force of gravity going down t1-w=ma m1: same thing ... t1-w=ma

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

<-m1 <-m2-> or <-m1-> <-m2->

Not sure but m1-> force coming from m2 and force going forward. m2->force from F (6.40N) and force of contact.

I get it. Thanks. Can you help me to draw a FBD for this problem ?

Here's the problem to make it clearer: 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...

Yes but here is what we did: 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 ?

Man I love physics. lol. Ok but why did we use m1 rather then m2 when we applied Newtons second law on m1 ?

To find the the answer in N ? 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...