Question

1.)

a.) A 29-kg crate rests on a horizontal floor, and a 58-kg person is standing on the crate.

(i) Determine the magnitude of the normal force that the floor exerts on the crate.
N

(ii) Determine the magnitude of the normal force that the crate exerts on the person.
N

b.) A Mercedes-Benz 300SL (m = 1600 kg) is parked on a road that rises 20° above the horizontal.

(i) What is the magnitude of the normal force?
N

(ii) What is the static frictional force that the ground exerts on the tires?
N

Answer 1

1) Let the mass of the crate is m = 29 kg

Let the mass of the person M = 58 kg

(i) Force due to person on the crate = Mg

Force due to crate on the floor is = mg

These two forces will be in downward direction

Let the normal force exerted by the floor on the crate be N

This normal force will be along upward direction

Since all the objects under consideration are at rest , total force acting on the system is zero

M g + m g - N = 0

0 = ( 58 * 9.8 ) + ( 29 * 9.8 ) - N

N = 568.4 + 284.2

N = 852.6 newton

Normal force exerted by the floor on the crate is equal to

N = 852.6 newton

(ii) From the above calculation , we know that the force exerted by the person on the crate is F = 568.4 N

Let N be the normal force exerted by the crate on the person

Since the crate and person are at rest , crate exerts a normal force of equal magnitude to this force on the person but in opposite direction.

So F - N = 0

F = N = 568.4 N

So normal force exerted by the crate on the person is N = 568.4 N

2) Let the mass of the car be m = 1600 kg

Angle at which the car is inclined is = 20°

The components of the gravitational force or weight of the car will be mg cos along the downward direction and mg sin along the horizontal direction

Since the car is at rest , all the horizontal and vertical forces acting on it must vanish separately.

(i) For this , let us consider the vertical forces

Forces acting along the vertical direction are , a normal force N in upward direction and vertical component of the weight of the car mg cos

So,. N - Mg cos = 0

N = mg cos

= 1600 * 9.8 * cos 20°

= 15680 * 0.93

= 14582.4 N

So the normal force acting on it is 14582.4 N

(ii) Horizontal forces acting on the car are , horizontal component of the weight of the car mg sin and a frictional force f equal and opposite to this force

This forces add to zero

f - mg sin = 0

f = mg sin

= 1600 * 9.8 * sin 20°

= 1600 * 9.8 * 0.34

= 5331.2 N

So static frictional force acting on it are f = 5331.2 N