When you take your 1300-kg car out for a spin, you go around a corner of radius 56.9 m with a speed of 15.8 m/s. The coefficient of static friction between the car and the road is 0.75. Assuming your car doesn't skid, what is the force exerted on it by static friction?
I got 5703.55 Kilonewtons but it's "off by a multiple of ten". Any help? Thank you so much.
A 16-foot uniform ladder rests against the side of a building. A 160 pound painter stands 4 feet from the top of the ladder. Determine the normal force acting on the ladder by the building and the frictional force acting on the ladder at the point it makes contact with the ground. ignore the weight of the ladder
divide by a 1000 your original answer is in N i believe also i would like to know how you solved this
To find the force exerted on the car by static friction, we need to consider the centripetal force acting on the car as it goes around the corner.
The centripetal force is given by the equation Fc = mv²/r, where Fc is the centripetal force, m is the mass of the car, v is its velocity, and r is the radius of the curve.
In this case, the mass of the car is 1300 kg, the velocity is 15.8 m/s, and the radius is 56.9 m. Substituting these values into the equation, we get:
Fc = (1300 kg) * (15.8 m/s)² / 56.9 m
= (1300 kg) * (249.64 m²/s²) / 56.9 m
≈ 5676.45 N
So the centripetal force acting on the car is approximately 5676.45 Newtons.
Now, to find the force exerted on the car by static friction, we need to compare this to the maximum possible static friction force, which is given by the equation Fsmax = μs * Fnorm, where Fsmax is the maximum static friction force, μs is the coefficient of static friction, and Fnorm is the normal force.
The normal force, in this case, is equal to the weight of the car, which can be calculated using the equation Fnorm = mg, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s²).
Fnorm = (1300 kg) * (9.8 m/s²)
= 12740 N
Substituting this value along with the coefficient of static friction (μs = 0.75) into the equation for the maximum static friction force, we get:
Fsmax = (0.75) * (12740 N)
= 9555 N
Therefore, the maximum static friction force that can be exerted on the car is 9555 Newtons. Since the centripetal force (5676.45 N) is less than the maximum static friction force, the car will not skid.
Hence, the force exerted on the car by static friction is approximately 5676.45 Newtons, which is equivalent to 5.67645 kilonewtons (kN).