Forces (1 Viewer)

[Lukybear]

A warehouse worker applies a force of 420N to push two cractes, (A,B) across the floor as shown in the figure 10.34. The friction force opposing the motion of the crates is a constant 2.0 N for each kilogram
A=30kg
B=40kg

>>>420N>>> AB
(where AB is side by side)

a)Calculate the acceleration of the crates
b)calculate the net force on the 40kg crate
c)calculate the force exerted by the 40 kg crate on the 30kg crate
d)calculate the force exerted by the 30kg crate on the 40kg crate
e) Would the worker find it easier to give the crates the same acceleration if the position of the two crates were reversed?

I really dont get the last 3. The answer says 240N. But I cannot get that... 260 N is what i got for c. Help!

 

[Lukybear]

Bump?

 

[k02033]

Let x be the force that box A applies to box B. Now by newton's 3rd law, what ever force box A applies to Box B, boxB will respond by applying force x back on to box A.
Let a be frictional force on box A and b be frictional force on boxB.
Take right to be the positive direction

so net force on boxA is 420-x-a.
so net force on boxB is x-b
so acceleration on boxA is 420-x-a/30 (as box A has mass of 30kg)
similarily acceleration on box B is x-b/40
now those 2 accelerations needs to be equal for the boxes to be moving together
so (420-x-a)/30=(x-b)/40
we know a=80N and b=60N
so now we can easily solve for x, which should be x=240N

notice force on B due to A is equal to force on A due to B but opposite directions

exchanging the positions of the box, will not make it easier, because just imagine the 2 boxes as one unit, in the end the guy has to move the same thing. (but the contact force between the boxes is different!) or another way is pushing the box from the left or the right makes no difference

 

[khorne]

This has been taken out of the syllabus, I believe.