A 600-kg car is going over a curve with a radius of 120 meters that is banked at an angle of 25 degrees with a speed of 30 meters per second. The coefficient of static friction between the car and the road is 0.3. What is the normal force exerted by the road on the car?
a) 7240 N
b) 1590 N
c) 5330 N
d) 3430 N
e) 3620 N
Note: The speed is not Vmax (roughly 32 m/s). I've been working under the assumption that mgcos(theta) will be increased by the frictional component acting downward. I simply can't seem to get the solution.
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A 600-kg car is going over a curve with a radius of 120 meters that is banked at an angle of 25 degrees with a speed of 30 meters per second. The coefficient of static friction between the car and the road is 0.3. What is the normal force exerted by the road on the car?
a) 7240 N
b) 1590 N
c) 5330 N
d) 3430 N
e) 3620 N
Note: The speed is not Vmax (roughly 32 m/s). I've been working under the assumption that mgcos(theta) will be increased by the frictional component acting downward. I simply can't seem to get the solution.
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http://i.imgur.com/kkaO7.png
the triangles that the dots/circles/particles are on are cross sections of the curve in the road.
the triangles coming from the dots/circles/particles are force vectors and their components parallel and perpendicular to the ramp.
Fn = normal force
Fc = centripetal force
Fg = force of gravity
Fgp = force of gravity's component perpendicular to the road
Fc = centripetal force's component perpendicular to the road
Fn = Fgp - Fcp
Fgp = cos(25)Fg
Fcp = -sin(25)Fc
Fg = mg -> (600kg)(9.81m/s/s)
Fc = (mv^2)/r -> (600kg)(30m/s)(30m/s)/(120m)
Fn = cos(25)(600kg)(9.81m/s/s) - sin(25)(600kg)(30m/s)(30m/s)/(120m)
Fn is approximately 3432.7N, so choice D.
the centripetal acceleration pulls the particle (car) towards the center of the circle. this decreases the normal force from the road, because the car is not pushing as hard into the ramp.
you can also use vector projection ( http://en.wikipedia.org/wiki/Vector_projection#Vector_projection_2 ) to project the centripetal acceleration along a unit vector perpendicular to the road.
to do this, you would define:
(note: ^ is denoting a unit vector, ~ is denoting a vector)
^r = the unit vector along the ramp
^p = the unit vector perpendicular to the ramp
~Fc = the vector of centripetal acceleration. points in -x direction.
Fc = magnitude of ~Fc, which is just (mv^2)/r
||~A|| = magnitude of vector A.
^r = (cos(25))^i + (sin(25))^j
^p = (sin(25))^i + (-cos(25))^j
~Fc = (-Fc)^i
projecting vector ~A along ~B:
( (~A dot ~B)/(||~B||^2) )*~B
projecting vector ~Fc along ^p
( (~Fc dot ^p)/(||^p||^2) )*^p
note: unit vectors are defined to have a magnitude of 1. therefore ||^p|| = 1
(~Fc dot ^p) = -sin(25)(Fc)
therefore the centripetal force subtracts a force from the net force "pushing" into the road.
the force of friction only acts tangential to the curve of the road, and therefore the dot product of this frictional force with the unit vector that is PERPENDICULAR to the ramp results in 0, meaning it does not effect the normal force.
The final answer is D.
@neugauss
This is not the right answer. Sorry, but I have no time right now to point exactly where your equations are incorrect.
This very question has already been answered on OS by viniterranova in the question by jgonzales8.
i'm struggling to see how my answer is incorrect. whenever you have the time to tell me what exactly is wrong with it, i'd love to hear.
viniterranova's solution is in broken english and it insists that the value of gravity is unknown. however, the idea is silly because gravity near the surface of the earth is constant??
i don't see how the gravity would be different at all?
and then he seems to solve for the "gravity" by assuming the answer to the question is 7240N? I don't see at all how he comes to that number?
please tell me where and how i'm incorrect, because the solution posted by viniterranova makes no sense to me at all.
Have a look at the drawing.
We have only 3 forces :
- W = mg weight, vertical downwards
- N normal force by the road, our unknown
- T friction force, parallel to the road, also unknown
|dw:1354741925183:dw|
N's 2nd law states that \(\vec W + \vec N + \vec T = m\vec a\)
As the motion is known (steady circular), the acceleration is horizontal and known:
a = v²/R
Now, as we want N and do not know the value of T, we project this relation on the axis normal to the road. It leads to:
\(-mg\cos \theta+N+0=m\Large\frac{v^2}{R}\normalsize \sin\theta\)
which leads to \(N=m(g\cos\theta+\Large\frac{v^2}{R}\normalsize \sin\theta)\)
with g = 9.8 or 10 m/s², the nearest answer is A.
ah yes, your answer makes more sense than viniterranova's answer, and our answers our essentially the same.
i disagree with the sign between your answer, though.
the normal force is from the road/ramp pushing back against the car with equal and opposite force that is being applied on it. the car will be applying some force on it -- toward the ramp -- and the centripetal acceleration will be applying some force as well -- AWAY from the ramp. the addition of the vector components of these two forces that are normal to the road/ramp will therefore be opposite in sign -- because one is pulling the car away from the ramp, and the other is pushing the car into the ramp.
therefore, the final answer must be
Fn = m[ cos(25)g - ( ((v^2)/R)sin(25) ) ]
which results in approximately 3430N, or choice D.
"and the centripetal acceleration will be applying some force as well -- AWAY from the ramp. "
This cannot be true. You are probably mistaking centripetal acceleration and centrifugal force.
Fn = m[ cos(25)g - ( ((v^2)/R)sin(25) ) ] is not possible because the faster the car, the greater Fn must be to help it change its velocity inwards.
Your formula, with the minus sign, says that the faster the car, the weaker the normal force. If the car is fast enough, Fn would become zero and the car would hover above the road: this is simply not possible.