# Newton's laws 2

by bumblebeeliz
Tags: laws, newton
 P: 38 1. The problem statement, all variables and given/known data Three blocks are in contact with each other on a frictionless horizontal surface as shown. The masses of the blocks are m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 3.0 kg. A horizontal force F = 24 N is applied to m1. a. Find the acceleration of the three blocks. b. Find the net force on each block. c. Find the magnitudes of the contact forces between the blocks. 2. Relevant equations Not sure.. $$\Sigma$$F = F1+F2+F3 3. The attempt at a solution a) F = a (1.0kg + 2.0 kg + 3.0 kg) 24N = a (6 kg) 24N / 6 kg = a 4 m/s2 = a b) F1 = 1kg (4 m/s2) = 4N F2 = 2kg (4 m/s2) = 8N F3 = 3kg (4 m/s2) = 12N c) F - F1 = 24N - 4N = 20N (contact force between F1 & F2) F - F1 - F2 = 24N - 4N - 8N = 12N (contact force between F2 & F3) ---- wow I feel lost after what I did.. Does this make sense? Attached Thumbnails
 P: 225 a) and b) look correct; however, contact forces isn't ringing a bell, so i can't say as much about c).
 P: 128 a) and b) are correct. As for contact force I presume the question is asking for the normal reaction, in which case c) is not correct. How would you calculate the normal reaction of the plane on the 1 kg, 2 kg and 3kg block?
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HW Helper
In part (c), the problem is asking for the contact force between each block, that is, what is the normal force (horizontal in this case) that block 1 exerts on block 2 (and vice versa, via Newton 3), and what is the normal (horizontal) contact force that block 2 exerts on block 3. So actualy, you have answered parts a, b, and c, correctly, BUT, your methodology is shaky at best. You should get used to drawing free body diagrams in problems such as this, noting all forces that act on the body and then using Newton's laws. In part (c), isolate Block 1 in a FBD. You know from Newton 2 that the sum of forces (the net force) acting on the body in the horizontal direction = ma. Thus, $$F - F_{21} = ma$$, where $$F_{21}$$ is the normal force of block 2 acting on block 1. Accordingly, $$24 - F_{21} = (1)(4)$$, from which, $$F_{21} = 20 N$$. And now isolate block 3, and the only force acting on it in the horizontal direction is the normal force of block 2 on block 3, thus $$F_{23} = ma$$, from which $$F_{23} = (3)(4) = 12 N$$. As a check, you can isolate block 2, or blocks 1 and 2 together, to achieve the same result. Again, your answers are correct, but I want to be sure you understand the reasoning behind the answers.