determine the average force on the box Do not round intermediate calculations; h

Question

determine the average force on the box

Do not round intermediate calculations; however, for display purposes report intermediate steps rounded to four significant figures. Give your final answer(s) to four significant figures. A 9 lb box is released from rest at a distance d = 5 ft from the bottom of a smooth chute with Theta = 24 degree. The friction between the chute and the box is negligible and so is the change in speed of the box in going from the inclined to the horizontal part of the chute. If the duration of the transition between inclined and horizontal motion takes 0.02 s, determine the average force acting on the box during this transition.

Explanation / Answer

From given figure,

We can find different force components

Now number of force acing on the body is,

So only two forces acts on body

Now downward force mg, resolves into two components

mgsin: In the direction of slope

mgcos: In the opposite direction of normal force.

So,

N- mgcos = 0

N =mgcos ………….equation 1

So,

Total force acting on the body is only mgsin

i.e. Fnet = mgsin………….equation 2

Put the value of m, g and sin, we will get average force.

Given values :

Mass m = 9 lb = 4.08 Kg

g= 9.8 m/s2

sin (24o) = 0.4067

So, equation 2 becomes,

Fnet = 4.08*9.8*0.4067 = 16.26 N = 3.65 lbs

The direction of force is,

F (avg)= -3.65 i – 3.65 j

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