CHAPTER 12. FORCE,MOMENTUM AND IMPULSE 12.3
SOLUTION
Step 1 : Draw a force diagram
Always draw a force diagram although the question might not ask forit. The
acceleration of the whole system is given, therefore a force diagram of the whole
system will be drawn. Because the two crates areseen as a unit, the forcediagram
will look like this:
15 kg
a = 2 m·s−^2
Applied force = 500 N
10 kg
Friction =?
Step 2 : Calculate the frictionalforce
To find the frictional force we will apply Newton’s Second Law. We are given
the mass (10 + 15 kg)and the acceleration (2m·s−^2 ). Choose the directionof
motion to be the positive direction (to the rightis positive).
FR= ma
Fapplied+ Ff= ma
500 + Ff= (10 + 15)(2)
Ff= 50− 500
Ff=− 450 N
The frictional force is 450 N opposite to the direction of motion (to the left).
Step 3 : Find the tension in therope
To find the tension in the rope we need to lookat one of the two crateson their
own. Let’s choose the 10 kg crate. Firstly, we need to draw a force diagram:
10 kg
(^13) of total frictional force Tension T
a = 2 m·s−^2
Ffon 10 kg crate
Figure 12.3: Force diagram of 10 kg crate
The frictional force on the 10 kg block is one third of the total, therefore:
Ff=^13 × 450
Ff= 150 N
If we apply Newton’s Second Law: