Find the contact force between a) boxes 1 and 2, and b) boxes 2 and 3.

No coefficient of friction supplied, or no friction? Which?

I’ll have to assume no friction.

Total mass = 9.4kg.

Acceleration = (f/m) = 7.5/9.4, = 0.798m/sec^2.

a) (3.2 + 4.9) = 8.1kg.

Force between 1 and 2 = (ma), = 8.1 x 0.798, = 6.464N.

b) Force between 2 and 3 = (ma), = 4.9 x 0.789, = 3.866N.

You did not supply an outline of ways those packing bins are prepared. they might properly be all stacked on accurate of one yet another. Your first answer assumed they were coated up from left to authentic contained in the order you listed them and the rigidity became pushing from the left. contained in the lines “rigidity between a million.3 and three.2 kg packing bins is 7.5 – a million.3*(0.8) = 6.40 six N”, the technique is 7.5 N became the unique rigidity, a million.3*(0.8) is the rigidity required to get the a million.3 kg container going with 0.8 m/s^2 acceleration. So the middle and authentic hand container have the last 6.40 six N. and particular adequate, F = (m2+m3)*a 6.40 six N = (3.20 kg+4.ninety kg)*0.8 m/s^2 … about, round off blunders.

So why is the contact between box 1 and 2 have the mass of 2 and 3?