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 4 ∘ angle with the vertical. Holding the ball in place takes a force of 20 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? Express your answer to two significant figures and include the

Answer 1

Answer:40.19 N

Explanation:

Given

Force needed to hold the ball at $\theta =4^{\circ}\ is\ F_1=20\ N$

From FBD we can write

$F\cos \theta =mg\sin \theta$

$\tan \theta =\dfrac{F}{mg}$

For $\theta =4^{\circ}$

$F_1=20\ N$

$\tan (4)=\dfrac{20}{mg}----1$

for $\theta =8^{\circ}$

Force is $F_2$

$\tan (8)=\dfrac{F_2}{mg}---2$

Divide 1 and 2 we get

$\dfrac{\tan (4)}{\tan (8)}=\dfrac{F_1}{F_2}$

$F_2=20\times \dfrac{\tan 8}{\tan 4}$

$F_2=20\times 2.009$

$F_2=40.19\ N$