You have three boxes. The 1.0 kg box is next to the 3.0 kg box, with the 2.0 kg

Question

You have three boxes. The 1.0 kg box is next to the 3.0 kg box, with the 2.0 kg box sitting on top of the 3.0 kg box. The 2.0 kg box docs not slip on the 3.0 kg box - they move together. Neglect friction between the boxes and the floor, and use g = 10 N/kg. Draw the free-body diagram of the three-box system, and use it to calculate the acceleration of the system Draw the free-body diagram of the 1.0 kg box, and use it to calculate the force the 3.0 kg box applies to the 1.0 kg box. Draw the free-body diagram of the 2.0 kg box, and use it to calculate the horizontal component of the force the 3.0 kg box applies to the 2.0 kg box. (This is a static friction force, but we really don't need to worry about what kind of force it is, at this point.) Draw the free-body diagram of the 3.0 kg box, and use it to calculate the normal force the floor applies to the 3.0 kg box. Also, figure out the net force acting on the 3.0 kg box two different ways and show that they are consistent with one another.

Explanation / Answer

a)
as there is no friction and the system moves together,

acceleration of the system of masses=24/(1+3+2)=4 m/s^2

force on the 1 kg box:

i)24 N , from left to right

ii)force by 3 kg box, from right to left

let force by 3 kg box=F N

then for 1 kg box,

24-F=1*acceleration

==> 24-F=1*4=4

==>F=24-4=20 N

forces on 3 kg box:

i) force due to 1 kg box, from left to right

ii) friction force with 2 kg box,, from right to left

let friction force be F1.

then for the 3 kg box:

20-F1=3*4

==>F1=20-12=8 N

c) force on the 2 kg box:

friction due to 3 kg box, from left to right

friction force is found from the previous stage as 8 N

and can be verified as 2*acceleration=8 N

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