The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F = 0.550mg is then applied at upward angle θ = 24°.
(a) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are μ_{s} = 0.610 and μ_{k} = 0.505?
(b) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are μ_{s} = 0.395 and μ_{k} = 0.295?
What is the vertical component and horizontal component of the applied force? What is the gravitational force? What is the vertical acceleration? From Newton's second law for vertical motion, what is the normal force? What is fs,max? How does that value compare with the horizontal component of the applied force?
(a) The magnitude of acceleration of the block across the floor which is given as :
from a free body diagram, we have
on the y-direction : F_{app} sin + F_{N} = F_{g}
F_{app} sin 24^{0} + F_{N} = mg
F_{N} = mg - (0.55 mg) sin 24^{0}
F_{N} = mg [1 - (0.55) sin 24^{0}] { eq.1 }
on the x-direction : F_{app} cos - F_{f} = F_{x}
F_{app} cos 24^{0} - _{k} F_{N} = m a { eq.2 }
(0.55 mg) cos 24^{0} - _{k} mg [1 - (0.55) sin 24^{0}] = m a
using, _{k} = 0.505
(because, the block is initially stationary and required a force greater than the maximum force of kinetic friction)
a = (0.55 g) cos 24^{0} - (0.505) g [1 - (0.55) sin 24^{0}]
a = (0.55) (9.8 m/s^{2}) (0.9135) - (0.505) (9.8 m/s^{2}) [1 - (0.55) (0.4067)]
a = (4.92 m/s^{2}) - (3.84 m/s^{2})
a = 1.08 m/s^{2}
(b) The magnitude of acceleration of the block across the floor which is given as :
F_{app} cos 24^{0} - _{s} F_{N} = m a
(0.55 mg) cos 24^{0} - _{s} mg [1 - (0.55) sin 24^{0}] = m a
using, _{s} = 0.395
(because, the block is initially stationary and required a force greater than the maximum force of static friction)
a = (0.55 g) cos 24^{0} - (0.395) g [1 - (0.55) sin 24^{0}]
a = (0.55) (9.8 m/s^{2}) (0.9135) - (0.395) (9.8 m/s^{2}) [1 - (0.55) (0.4067)]
a = (4.92 m/s^{2}) - (3 m/s^{2})
a = 1.92 m/s^{2}
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