A m=10 kg box sits on a level surface where the coefficient of kinetic friction

Question

A m=10 kg box sits on a level surface where the coefficient of kinetic friction is µk = 0.5. Bob comes along and pushes the box forward at a constant speed by pushing on the box with a force FBob at an angle of 30 degree below the horizon. Thomas comes along and pulls the box forward at a constant speed by pulling on the box with a force FTom at an angle of 30 degree above the horizon. (a) Find the normal force for box for Bob’s and Thomas’ situation in terms of FBob and FTom. (b) Find the frictional force in each case (again in terms of FBob and FTom. (c) Noting that the boxes move at constant speed on level ground, find FBob and FTom. Show that Thomas exerts less force than Bob.

Explanation / Answer

(a) For Bob's situation:

Nbob = mg + Fbob(sin 30o) = 10*9.8 + Fbob/2 = 98 + Fbob/2

For Tom's situation:

Ntom = mg - Ftom(sin 30o) = 10*9.8 - Fbob/2 = 98 - Ftom/2

(b) For Bob's situation:

fbob = k*Nbob = 0.5(98 + Fbob/2) = 49 + Fbob/4

For Tom's situation:

ftom = k*Ntom = 0.5(98 - Ftom/2) = 49 - Ftom/4

(c) Since horizontal acceleration is zero, horizontal forces cancel each other out.

For Bob's case:

Fbob(cos 30o) = fbob = 49 + Fbob/4

=> Fbob = 49/(cos30o - 0.25) = 79.5 N

For Tom's case:

Ftom(cos 30o) = ftom = 49 - Ftom/4

=> Ftom = 49/(cos30o + 0.25) = 43.9 N

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