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You are designing a system for moving aluminum cylinders from the ground to a loading dock. You use a sturdy wooden ramp that is 9.00 m long and inclined at 37.0∘ above the horizontal. Each cylinder is fitted with a light, frictionless yoke through its center, and a light (but strong) rope is attached to the yoke. Each cylinder is uniform and has mass 470 kg and radius 0.300 m. The cylinders are pulled up the ramp by applying a constant force F to the free end of the rope. F is parallel to the surface of the ramp and exerts no torque on the cylinder. The coefficient of static friction between the ramp surface and the cylinder is 0.120.

a) What is the largest magnitude F can have so that the cylinder still rolls without slipping as it moves up the ramp?
b) If the cylinder starts from rest at the bottom of the ramp and rolls without slipping as it moves up the ramp, what is the shortest time it can take the cylinder to reach the top of the ramp?

Respuesta :

akindeleot

Answer:

a = αR

Then We apply the Newton's second law of motion:

∑ F = ma

F - mg sinθ - f = ma.

F - mg sinθ - μmg cosθ = ma

Using given data we have:

F - 3217 = 470a ........................... (1)

Apply the Newton's law for rotation:

∑τ = I α

FR - (mg sinθ)R = [tex](\frac{mR^2}{2}+mR^2)\frac{a}{R}[/tex]

then: F - mg sinθ = 3ma / 2.

F - 2774 = 705a .......................... (2)

solving 1 and 2 we get:

F = 4103 N

a = 1.885 m/s²

b) In this case the time is given by:

t = [tex]\sqrt{\frac{2d}{a}}[/tex] it comes from: d = [tex]v_ot+\frac{1}{2}at^2[/tex] starting from rest vo = 0

t = [tex]\sqrt{\frac{2*9}{1.885}[/tex] = 3.09 s

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