1. What frictional coefficient is needed to keep a car moving at 86km/h on a 130 m -radius unbanked turn?
2. The handle of a 25kg lawnmower makes a 36 degree angle with the horizontal.
If the coefficient of friction between lawnmower and ground is 0.68, what magnitude of force, applied in the direction of the handle, is required to push the mower at constant velocity? (The answer is 410N).
Compare with the mower’s weight. (<-- this is what I am stuck on)
2. The handle of a 25kg lawnmower makes a 36 degree angle with the horizontal.
If the coefficient of friction between lawnmower and ground is 0.68, what magnitude of force, applied in the direction of the handle, is required to push the mower at constant velocity? (The answer is 410N).
Compare with the mower’s weight. (<-- this is what I am stuck on)
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1.
V = 86/3.6 = 23.89 m/s
Ar = V²/R = 4.39 m/s²
µ = Ar/g = 0.448
2.
F = µmg/[cosΘ - µsinΘ] = 407 N
F/mg = 407/(25*9.8) = 1.66
V = 86/3.6 = 23.89 m/s
Ar = V²/R = 4.39 m/s²
µ = Ar/g = 0.448
2.
F = µmg/[cosΘ - µsinΘ] = 407 N
F/mg = 407/(25*9.8) = 1.66
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1. What frictional coefficient is needed to keep a car moving at 86km/h on a 130 m -radius unbanked turn?
Convert the velocity to m/s
86 km/hr * 1000 m/km * 1 hr/3600 s = 23.889 m/s
Since the turn is unbanked, the friction force must equal the centripetal force.
Friction force = µ * m * g
Centripetal force= m * v^2 ÷ r
µ * m * g = m * v^2 ÷ r
Divide both sides by m * g
µ = v^2 ÷ (r * g)
2. The handle of a 25kg lawnmower makes a 36 degree angle with the horizontal.
If the coefficient of friction between lawnmower and ground is 0.68, what magnitude of force, applied in the direction of the handle, is required to push the mower at constant velocity? (The answer is 410N).
Compare with the mower’s weight. (<-- this is what I
http://s3.amazonaws.com/answer-board-ima…
The website above has a picture, which is helpful.
The force, applied in the direction of the handle, has vertical and horizontal component.
Vertical component = F * sin 36˚
Horizontal component = F * cos 36˚
Friction force = µ * Total force perpendicular to the ground.
The weight of the mower and the vertical component of the force applied in the direction of the handle are perpendicular to the ground.
Weight = 25 * 9.8 = 245 N
Friction force = 0.68 * (F * sin 36˚ + 245)
To push the mower at constant velocity the horizontal component of the force applied in the direction of the handle are perpendicular to the ground must equal the friction force.
F * cos 36˚ = 0.68 * (F * sin 36˚ + 245)
Solve for F
F * cos 36˚ = (0.68 * F * sin 36˚) + (0.68 * 245)
F * cos 36˚ – (0.68 * F * sin 36˚) = (0.68 * 245)
F * (cos 36˚ – 0.68 * sin 36˚) = (0.68 * 245)
F = 407 N
Weight = 245 N
The force, applied in the direction of the handle, is greater than weight of the mower.
Convert the velocity to m/s
86 km/hr * 1000 m/km * 1 hr/3600 s = 23.889 m/s
Since the turn is unbanked, the friction force must equal the centripetal force.
Friction force = µ * m * g
Centripetal force= m * v^2 ÷ r
µ * m * g = m * v^2 ÷ r
Divide both sides by m * g
µ = v^2 ÷ (r * g)
2. The handle of a 25kg lawnmower makes a 36 degree angle with the horizontal.
If the coefficient of friction between lawnmower and ground is 0.68, what magnitude of force, applied in the direction of the handle, is required to push the mower at constant velocity? (The answer is 410N).
Compare with the mower’s weight. (<-- this is what I
http://s3.amazonaws.com/answer-board-ima…
The website above has a picture, which is helpful.
The force, applied in the direction of the handle, has vertical and horizontal component.
Vertical component = F * sin 36˚
Horizontal component = F * cos 36˚
Friction force = µ * Total force perpendicular to the ground.
The weight of the mower and the vertical component of the force applied in the direction of the handle are perpendicular to the ground.
Weight = 25 * 9.8 = 245 N
Friction force = 0.68 * (F * sin 36˚ + 245)
To push the mower at constant velocity the horizontal component of the force applied in the direction of the handle are perpendicular to the ground must equal the friction force.
F * cos 36˚ = 0.68 * (F * sin 36˚ + 245)
Solve for F
F * cos 36˚ = (0.68 * F * sin 36˚) + (0.68 * 245)
F * cos 36˚ – (0.68 * F * sin 36˚) = (0.68 * 245)
F * (cos 36˚ – 0.68 * sin 36˚) = (0.68 * 245)
F = 407 N
Weight = 245 N
The force, applied in the direction of the handle, is greater than weight of the mower.