Question

A heavy steel ball is hung from a cord to make a pendulum. The ball is pulled to the side so that the cord makes a 5 ∘ angle with the vertical. Holding the ball in place takes a force of 40 N . If the ball is pulled farther to the side so that the cord makes a 8 ∘ angle, what force is required to hold the ball?

Answer 1

64 N

Step-by-step explanation:

Given that the angles are very small, the following approximation can be made:

F ≈ m*g*α

where F is the force needed to hold the ball, m is the ball mass, g is the acceleration of gravity and α is the angle between the cord and the vertical.

Let's call F1 the force needed to hold the ball at 5° (α1) and F2 the force needed to hold the ball at 8° (α2).

F1 ≈ m*g*α1

F2 ≈ m*g*α2

Dividing the equations:

F1/F2 = α1/α2

F2 = (α2/α1)*F1

F2 = (8/5)*40

F2 = 64 N

Answer 2

f = 63.8 N

Step-by-step explanation:

initial angle to the vertical = 5 degrees

initial holding force = 40 N

final angle to the vertical = 8 degrees

final holding force = ?

find the final holding force

• force = mgSinθ

m = mass and g = acceleration due to gravity

• for the initial holding force:

40 = mgSin5

mg =
`(40)/(sin5)` ....equation 1

• for the final holding force:

f = mgSin8 ......equation 2

substituting the value of mg from equation 1 (where mg =
`(40)/(sin5)` ) into equation 2

f = mgSin8 =
`(40)/(sin5)` x Sin8

f = 63.8 N