Posted by Cesar on .
Blocks of mass 4, 8, and 24 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 15 N is applied to the left-most block.
a.)What is the magnitude of the force that the middle block exerts on the rightmost one?
b.)What is the magnitude of the force that the leftmost block exerts on the middle one?
c.)Suppose now that the left-right order of the blocks is reversed. Now find the magnitude of the force that the leftmost block exerts on the middle one?
a) Keep applying F = ma. First you need the total acceleration, a. Get it by applying F = ma to all three blocks, using the externally applied F = 15 N
a = F/m = 15 N/36 kg = 0.4167 m/s^2
Then apply it again, for the rightmost block only.
F'' = 24 a = 10 N
b) F' = 32 a = 13.33 N
c) F = 12 a = 5 N