### Newton : Tut 2

**1.** The diagram shows a truck pulling two trailers with an applied force of 30N.

**1.1.** Draw each trailer showing and labelling all the forces acting on them.

1.1.

**1.2.** Calculate the acceleration of the system.

1.2. The acceleration of any of the objects would be the acceleration of the system.

**1.3.** Calculate the coefficient of kinetic friction of the 4kg trailer.

1.3.

**1.4.** Calculate the friction of the 3kg trailer.

1.4.

**1.5.** Calculate the coefficient of friction for the 3kg trailer.

1.5.

**2.** Consider the diagram. A force of 30N is pulling two trolleys. The tension in the connecting string is unknown at T.

**2.1.** Calculate the friction force experienced by the 6kg trolley.

2.1.

**2.2.** Calculate the magnitude of the acceleration of the system.

2.2.

**2.3.** Calculate the magnitude of the tension T.

2.3.

**3.** A force of 46N is applied horizontally on a 2kg box as shown. The 2kg box in turn, pushes against an 8kg box. The individual friction forces are shown.

**3.1.** Draw a force diagram for each box, showing all the horizontal forces. Label.

3.1.

F_{C} represents the Contact Force between the two boxes. As a result of N(III), each box would exert the same size force on the other.
**3.2.** Calculate the magnitude of the Contact Force.

3.2.

**4.** A 2kg block is placed on top of a 10kg trolley. The block is pulled with a force of 20N. The friction force
between the block and the top surface of the trolley is then 6N. The trolley itself experiences no significant friction force with the ground.

**4.1.** Draw a force diagram for the 2kg block, showing all the horizontal forces acting on it. Hence calculate the acceleration of the block.

4.1.

**4.2.** Draw a force diagram for the 10kg trolley showing all the horizontal forces acting on it. Hence calculate the acceleration of the trolley.

4.2.

- The 6N friction is common to both objects.
- The 6N friction opposes motion for the block.
- The 6N actually makes the 10kg trolley move.
- Although the 2kg block is on top of the 10kg trolley, these are two separate masses and must be treated as such.

Therefore in the N(II) equation in 4.2., the mass used for the trolley is just 10kg and NOT 12kg.
- The original 20N force DOES NOT act on the 10kg trolley. Only on the 2kg block.

Therefore the 20N force is not drawn on the trolley.