To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal
force and reduce vehicle traction it is important to
1 answer:
Answer: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
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Answer
(C).
When there is an angle between the two directions, the cosine of the angle must be considered.
Step by step Solution
The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.
The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.
The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is .
Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the meaning that there is no work done in the direction perpendicular to the motion.
The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is