You have three boxes The 1.0 kg box is next to the .1.0 kg box, with the 2 0 kg

Question

You have three boxes The 1.0 kg box is next to the .1.0 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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