Question 467.
Imagine you are dragging your sibling on a sled across a flat, snowy surface at a constant speed by hauling on a rope attached to the front of the sled. The rope makes an angle θ with respect to the ground. If the coefficient of kinetic friction μk between the sled and the snow is 0.1, then what should θ be in radians if you want to exert the least amount of force necessary to keep the sled going?
Answer 467.
Let F = force applied
and R = normal reaction from the ground
=> Fsinθ + R = mg
=> R = mg - Fsinθ
Horizontal component of the force balances the kinetic frictional force for the sled to move with constant velocity
=> Fcosθ = (mg - Fsinθ) * 0.1
=> F (cosθ + 0.1 sinθ) = 0.1 mg
F will be minimum when cosθ + 0.1 sinθ is maximum
cosθ + 0.1 sinθ
= r [(1/r) cosθ + (0.1/r) sinθ], where r = √[1 + (0.1)^2]
= r [cos(α - θ)], where cosα = 1/r
Maximum value of cosθ + 0.1 sinθ will be for
θ = α
= arccos(1/r)
= arccos 1/√[1 + (0.1)^2]
= 5.71°.
and R = normal reaction from the ground
=> Fsinθ + R = mg
=> R = mg - Fsinθ
Horizontal component of the force balances the kinetic frictional force for the sled to move with constant velocity
=> Fcosθ = (mg - Fsinθ) * 0.1
=> F (cosθ + 0.1 sinθ) = 0.1 mg
F will be minimum when cosθ + 0.1 sinθ is maximum
cosθ + 0.1 sinθ
= r [(1/r) cosθ + (0.1/r) sinθ], where r = √[1 + (0.1)^2]
= r [cos(α - θ)], where cosα = 1/r
Maximum value of cosθ + 0.1 sinθ will be for
θ = α
= arccos(1/r)
= arccos 1/√[1 + (0.1)^2]
= 5.71°.
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