Physics help 1 A car is traveling up a hill that is inclined

Physics help

1) A car is traveling up a hill that is inclined at an angle above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a)   = 14^o and (b)   = 38^o

2) A car (m = 1590 kg) is parked on a road that rises 16° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

3) While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.370. The pushing force is directed downward at an angle below the horizontal. When is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find that value of .

(a) Ratio Number (b) Ratio Number Units Units

Solution

1)

The weight acts vertically downward and is given by (mg)

The normal force is given by [mg cos(theta)]

So the ratio of normal to weight is

mg*cos(theta) / mg = cos (theta)

a) Ratio = cos 14 = 0.97

b) Ratio = cos 38 = 0.788

2)

a)

You need to work out the forces perpendicular to the road. So we can say:

Fn = mgcos(theta)

Fn = (1590 x 9.81 x sin(16))

Fn = 4299.36 N

This is the normal force acting against the car.

b)

We can say that since the car is not moving, something is stopping the car from moving forwards. The friction is

doing this. We can summarise that the force of the friction is acting parallel to the road.

Fs = (1590 x 9.81 x cos(16)) = 14993.66 N