(a) A 10 kg box of candy rests on a floor with a coefficient of static friction of 0.30. What force is needed to move the box?
b) What is the coefficient of kinetic friction between a rubber tire and the road if a 2000 kg car needs 1.57 × 104 N to keep the car moving at a constant speed?
normal force = m g = 10*9.81
so
friction force max = .3 * 10 * 9.81
which is min horizontal force required to start acceleration
F = m a
a = 0
so
friction Force - pulling force = 0
mu * 2000 * 9.81 = 1.57 * 10^4
(a) To find the force needed to move the box, we can use the formula for static friction:
Fs = μs * N
Where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.
Given:
Mass of the box (m) = 10 kg
Coefficient of static friction (μs) = 0.30
The normal force N is equal to the weight of the box:
N = m * g
Where g is the acceleration due to gravity, approximately 9.8 m/s^2.
N = 10 kg * 9.8 m/s^2 = 98 N
Now we can calculate the force of static friction:
Fs = 0.30 * 98 N = 29.4 N
Therefore, a force of 29.4 Newtons is needed to move the box.
(b) To find the coefficient of kinetic friction, we can use the formula for kinetic friction:
Fk = μk * N
Where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
Given:
Mass of the car (m) = 2000 kg
Force required to keep the car moving (Fk) = 1.57 × 10^4 N
The normal force N is again equal to the weight of the car:
N = m * g
N = 2000 kg * 9.8 m/s^2 = 19600 N
Now we can calculate the coefficient of kinetic friction:
μk = Fk / N
μk = (1.57 × 10^4 N) / (19600 N)
μk ≈ 0.80
Therefore, the coefficient of kinetic friction between the rubber tire and the road is approximately 0.80.
To answer both of these questions, we need to understand the concept of friction.
(a) The force needed to move the box can be determined using the formula:
Force of friction (Ff) = coefficient of static friction (μs) * Normal force (Fn)
The normal force (Fn) is the force exerted by the surface perpendicular to the object resting on it, which is equal to the weight of the object. In this case, the weight is the mass (m) of the box multiplied by the acceleration due to gravity (g).
So, the normal force (Fn) = m * g = 10 kg * 9.8 m/s^2 = 98 N.
Given the coefficient of static friction (μs) = 0.30, we can substitute these values into the formula:
Ff = μs * Fn = 0.30 * 98 N = 29.4 N
Therefore, the force needed to move the box is 29.4 Newtons.
(b) The coefficient of kinetic friction (μk) can be calculated using the formula:
Force of friction (Ff) = coefficient of kinetic friction (μk) * Normal force (Fn)
In this case, the normal force (Fn) is the weight of the car, which is equal to the mass (m) multiplied by the acceleration due to gravity (g).
So, the normal force (Fn) = m * g = 2000 kg * 9.8 m/s^2 = 19,600 N.
Given that the force required to keep the car moving at a constant speed is 1.57 × 10^4 N, we can substitute the values into the formula:
Ff = μk * Fn
1.57 × 10^4 N = μk * 19,600 N
Dividing both sides by 19,600 N, we get:
μk = (1.57 × 10^4 N) / (19,600 N) ≈ 0.80
Therefore, the coefficient of kinetic friction between the rubber tire and the road is approximately 0.80.