Engineers are designing a curved section of a highway. If the radius of curvature of the curve is 173 m, at what angle should th
e curve be banked so that a car traveling at 29.0 m/s will stay on the road without the aid of frictional forces?
1 answer:
Answer: The angle Ø = 26
Explanation:
The weight and the normal force of the road are the only two external forces acting on the car. A frictionless surface can only exert a force that is perpendicular to the surface. This is a normal force.
Since the car does not leave the surface of the road, the vertical components of the two external forces must be equal in magnitude and opposite in direction. Whereas ;
Horizontal component = centripetal force. While
The vertical component = Mg
Revolving the two components gives: tanØ = V^2/rg
Please find the attached file for the solution
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Answer:
b. mg ( μ · cos θ + sin θ)
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following parallel forces acting on the block (in parallel direction to the direction of movement):
F = applied force.
Fr = friction force.
wx = parallel component of the weight.
According to Newton´s second law:
∑F = m · a
Where "m" is the mass of the block and "a" its acceleration.
Then:
F - Fr - wx = m · a
Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:
F - Fr - wx = 0
F = Fr + wx
The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.
The friction force is calculated as follows:
Fr = μ · N
Where N is the normal force and μ is the coefficient of friction.
Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.
Let´s apply Newton´s second law in the perpendicular direction to show this:
∑F = m · a
N - wy = m · a
The acceleration of the block in the perpendicular direction is zero. Then:
N - wy = 0
N = wy
And wy can be obtained by trigonometry (see figure):
wy = W · cos θ
N = wy = mg · cos θ
The parallel component of the weight is calculated using trigonometry (see figure):
wx = W · sin θ
wx = mg · sin θ
Then the applied force will be:
F = Fr + wx
F = μ · N + mg · sin θ (N = wy = mg · cos θ)
F = μ · mg · cos θ + mg · sin θ
F = mg ( μ · cos θ + sin θ)
The correct answer is the b.
