Forces Physics problem?
A 5.90-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.420. Determine the kinetic frictional force that acts on the box when the elevator is
(a) stationary,
(b) accelerating upward with an acceleration whose magnitude is 1.50 m/s2, and
(c) accelerating downward with an acceleration whose magnitude is 1.50 m/s2.
Comments
a) F = 5.9 x 9.8 x 0.420 = 24.2844 N
b) F = 5.9 x ( 9.8 - 1.5 ) x 0.420 = 20.5674 N
c) F = 5.9 x ( 9.8 + 1.5 ) x 0.420 = 28 N
Goodbye
Sum Fx = zero and Sum Fy = zero Sum Fx = F1,x + F2,x + F3,x = zero + forty four cos 60 + F3,x = 22 + F3,x = zero. So, F3,x = -22 N. Sum Fy = F1,y + F2,y + F3,y = 33 + forty four sin 60 + F3,y = 33 + forty four sin 60 + F3,y = zero So, F3,y = -seventy one N. F3 = sqrt [ F3,x ^two + F3,y ^two ] = sqrt [ (-22)^two + (-seventy one)^two ] = seventy four N. Direction of F3 = a hundred and eighty + tan^(-a million) (seventy one/22) = a hundred and eighty + seventy three = 253 deg.