Ok, I assume you didn't mean side to side, but they are in a front to back line, 1 in front, 3 in back, and you are pushing on 3.
find the net acceleration
F=ma
7.5=(summasses)a solve for a.
Now
1) box one is being pushed, so force=m1*a
2) box 2 and 1 are being pushed, so
f=(M1+m2)a
A force of magnitude 7.50 N pushes three boxes with masses m_1 = 1.30 kg,m_2 = 3.20 kg, and m_3 = 4.90 kg .The mass is side by side.
1-Find the magnitude of the contact force between boxes 1 and 2, and
2-Find the magnitude of the contact force between boxes 2 and 3.
*the position of the box is like this:-
7.50N(push)-> |m1|m2|m3|
1 answer
1 answer