Physics Help Please?

Question

When you take your 1200 kg car out for a spin, you go around a corner of radius 58.1 m with a speed of 15.8 m/s. The coefficient of static friction between the car and the road is 0.69. Assuming your car doesn't skid, what is the force exerted on it by static friction?

Please help. Neither I nor my study group can figure this one out.

Answer 1

Well you know that F=ma

You know the mass of the car.

Going around a turn at a constant speed means that your acceleration is centripetal acceleration.... v^2 / r

So the force exerted on the car by the road... or the force between the road and the tires will be equal to mv^2 / r

To find out of the car skids or not you have to find the maximum value of friction which must be larger than the centripetal force. The maximum value of friction will just be .69 times the mass of the car because gravity and normal force are the only forces acting in the y direction.

Answer 2

First, ID the physics of the problem. In this case, we are looking at a force issue; we can tell that from the words "static friction" where friction force f = kN = k mg cos(theta); and k = .69, m = 1200 kg, g = 9.81 m/sec^2 on Earth's surface, and theta is presumed zero degrees incline of the road (that is we presume the road is flat in the curve).

Then, because this is a force physics problem, we line up what we know about forces in this context. Newton's stuff is germane when talking about motion and forces. And prominent among his so-called laws is F = Ma.

Because the car is not skidding, we know from the laws of physics that acceleration of the car in the direction of the radius of turn R = 58.1 m is a = 0. Why? Because there is no net force acting in that direction since the radial velocity V(along R) = 0 mps, a constant. And when a velocity is constant, there is no acceleration because dV/dt = 0 = a.

OK, then, when a car is not skidding, F = ma = 0 = (C - f); where centrifugal force C = mv^2/R and v = 15.8 mps the trangential velocity of the car around R. As there is no net force, C and f are equal but opposite forces and cancel each other out. Thus we have C = mv^2/R = kmg = f.

And there you have it...the static friction f = mv^2/R and m, v, and R are given. You can do the math.

But let's review the physics. First, this is a force problem. Second, when a velocity like the radial velocity is zero, that's a constant velocity. Third, when velocity is constant, there is no acceleration; so net forces in the direction of the velocity (e.g., along R) is zero from F = ma. Finally, when F = 0, all the forces acting on the mass m cancel out to give that zero net value.

Answer 3

To go round a curve there must be a centripetal force of value
mv^2/r where m is the mass v is the velocity and r is the radius of the curve.

The frictional force which acts toward the center of the curve is the centripetal force.
The maximum frictional force is μ mg.

But if the speed is less then the frictional force will be less and will be equal to mv^2 /r.
The maximum frictional force = 0.69 x 1200 x 9.8 = 8114.4 N.

Let us calculate the actual frictional force by
mv^2 /r = 1200 x 15.8 ^2 / 58.1 = 5156N.
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Since this frictional force is less than the maximum frictional force of 8114 N the car will not slide.

However, if the speed increases , then the frictional force will also increase but up to the maximum of 8114 N.
We can find the corresponding maximum speed using the formula mv^2 /r.
8114 = 1200 v^2 / 58.1
v = 19.82 m/s.
If the speed increases more than this value then the frictional force will not increase and hence the car will skid .