forces and newtonian physics help please?

Question

A 1300kg car pulls a 540kg trailer away from a stop light with an acceleration of 1.92 m/s2

a. What is the net force exerted by the car on the trailer in N?
b. What is the magnitude of the force that the trailer exerts on the car in N?
(c) What is the net force acting on the car in N?

A force of magnitude 8.11 N pushes 3 boxes with masses m1 = 1.30 kg, m2 = 3.20kg, and m3 = 4.90kg. Find the contact force between boxes 1 & 2, and between boxes 2 & 3.

Two crewmen pull a boat through a lock. One crewman pulls with a force of 130 N at an angle, θ = 35°, relative to the forward direction of the boat. The 2nd crewman, on the opp side of the lock, pulls at an angle of 45° degrees relative to the forward direction. With what force should the second crewman pull so that the net force of the two crewmen is in the forward direction?

A train is traveling up a 3.2 degree incline at 3.74 m/s when the last car breaks free starts to coast. How long does it take for the last car to come to rest?

Answer 1

The first one is fairly tricky. We have a situation where the car's engine must exert enough force to move the car and the trailer and a certain rate of acceleration. The car must exert a net force on the trailer to accelerate it. That net force will be equal to:

F = m * a = 540 kg * 1.92 m/s2 = 1040 N

By Newton's third law we know if the car exerts a force of 1040 N on the boat, the boat must exert a force of 1040 N on the car in the opposite direction.

This force added to the force to accelerate the mass of the car will give us the total force which must be exerted by the car.

1040 N + 1300 kg * 1.92 m/s2 = 3530 N



For your second question, the acceleration of all three boxes MUST be equal. This would imply that each box feels a different net force. To find this acceleration we add the masses of all three boxes and solve for acceleration.

F = (m1+ m2+ m3) a => a = F/(m1+m2+m3) = 8.11 N /(1.3 kg+ 3.2 kg + 4.9 kg) = 0.862 m/s2

If the force is acting on box one from the left and box 2 is to the right of box 1 and box 3 is to the right of box 2, we can find the contact forces. The force needed to accelerate box three is 4.23 N, this must be the contact force between 2 and 3. The contact force between 1 and 2 is the force needed to accelerate both box 2 and box 3 or 6.99 N.

Answer 2

a)
The trailer is pulled with an acceleration of 1.92m/s^2 implies both the car and trailer moves with an acceleration of 1.92m/s^2.

Total mass is 1300 + 540 = 1840 kg.

Net force is mass x acceleration = 1840 x 1.92 = 3532.8N.

b)

The trailer of mass 540kg moves with an acceleration of 1.92m/s^2.

The force acting on the trailer is 540 x 1.92 = 1036.8 N

c)

The car moves with an acceleration of 1.92 m/s^2.

The force that must act on this is 1300 x 1.92 = 2496 N

2.

The three boxes must have the same acceleration to have a constant contact.

8.11N force pushes all the three boxes of total mass 9.4 kg
a = Force / mass =8.11 / 9.4 = 0.863 m/s^2.

Box 1 pushes with force on the boxes of masses (3.2 +4.9 = 8.1) with an acceleration of 0.863 m/s^2.
Force on 2 and 3 is 8.1* 0.863 = 6.99 N.

Boxes 1 and 2 pushes the box 3 of mass 4.9 with an acceleration of 0.863 m/s^2.
Force on 3 is 4.9* 0.863 = 4.23 N.

3.)

F sin 45 = 130 sin 35

F = 105.45 N.

4)
Acceleration along the plane = 9.8 sin 3.2 = 0.547m/s^2.

a = v-u / t
where v =0 and u = 3.74 m/s and a -0.547 m/s^2.

t = - 3.74 /-0.547 =6.84 s.