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

Answer:

f = 63.8 N

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 = $\frac{40}{sin5}$ ....equation 1

• for the final holding force:

        f = mgSin8  ......equation 2

   substituting the value of mg from equation 1 (where mg = $\frac{40}{sin5}$ ) into equation 2

    f = mgSin8 = $\frac{40}{sin5}$ x Sin8

   f = 63.8 N

       

Answer 2

Answer:

64 N

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