Everything Science Grade 11

CHAPTER 12. FORCE,MOMENTUM AND IMPULSE 12.3

20 kg

60 ◦

150 N

15 N

Fx

Step 2 : Calculate the horizontal component of the applied force

The applied force is acting at an angle of 60◦to the horizontal. We can only

consider forces that areparallel to the motion. The horizontal component of the

applied force needs to be calculated before we can continue:

Fx = 150 cos 60◦

Fx = 75N

Step 3 : Calculate the acceleration

To find the accelerationwe apply Newton’s Second Law:

FR = ma

Fx+ Ff = (20)(a)

75 + (−15) = 20a

a = 3m· s−^2 to the right

Example 8: Newton II - Truck andtrailer

QUESTION

A 2000 kg truck pulls a500 kg trailer with a constant acceleration. The engine of the truck

produces a thrust of 10000 N. Ignore the effectof friction.

1. Calculate the acceleration of the truck.


2. Calculate the tensionin the tow bar T between the truck and the trailer, if the tow bar
   makes an angle of 25◦with the horizontal.

25 ◦

a =? m·s−^2

500 kg 2000 kg

10 000 N

T

Figure 12.4: Truck pulling a trailer