The net force is along the horizontal floor, since it does not leave the floor vertically. It equals the horizontal component of the pulling force minus the friction force. The friction force is somewhat reduced by the vertical component of the pulling force (250*sin32).
Fnet = 250 cos32 - 0.18(M*g-250*sin32)
= 212 - [0.18(941 - 132.5)]
= 212 - 145.5
= 66.5 N
A 250N force is applied at an angle of 32 degrees above the horizontal to a 96kg box causing it to slide along a floor. The coefficient of friction between the box and the floor is 0.18. What is the magnitude of the net force on the box? The answer is 67N, I'm just not sure how to get there.
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