Lecture 7: Forces

Example 1: A firefighter dashes up a 26 m long ladder, making an angle of 60 degrees with the ground.

• What is the vertical displacement of her feet at the top? • What is the horizontal displacement of her feet at the top?

Look at the triangle formed by the length of the ladder 𝑠 = 26  m, the vertical displacement 𝑠! (the position of the top of the ladder), and the horizontal displacement 𝑠! (the position of the foot of the ladder). Using the sine rule, we have

sin 60° =𝑠!26  m

so 𝑠! = 26 sin 60° = 26×0.866 = 22.5  m

Similarly, using the cosine rule,

cos 60° =𝑠!26  m

so 𝑠! = 26 cos 60° = 26×0.5 = 13  m

Example 2: An aircraft in a 45 degree dive is travelling at a constant 800 km/h. Determine its speed in m/s in the vertical direction – i.e. the rate at which the plane is losing altitude. Solution: The y-component of the plane’s velocity is given by

sin 45° = !!!

so

𝑣! = 𝑣 sin 45° = 800  km/h×0.707= 566  km/h

vertically downwards.

Example 3: Determine the resultant (i.e. net force) exerted on the stationary elephant by the two clowns. What is the magnitude and direction of the force? Solution: The net force exerted on the elephant is given by the sum of the two forces exerted by the clowns 𝐹! = 300  N due north, and 𝐹! = 400  N due east, so the sum of these two is given by the following triangle, with sides 𝐹!, 𝐹! and 𝐹.

Since the triangle is a right-angled triangle, we can find the length of the hypotenuse using Pythagoras:

𝐹! = 𝐹!! + 𝐹!! = 300! + 400!= 900+ 1600 = 2500

so 𝐹 = 2500 = 500  N

We can find the direction of the force using

tan𝜃 =400300 =

43

so

𝜃 = tan!!43 = 53°

so the net force on the elephant is 53° east of north. Since the elephant is balancing this pull, and is stationary, the elephant is pulling with force –𝐹, so with a force of 500 N in a direction 53° west of south.