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

Step-by-step 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 =(F)/(mg)`

For
`\theta =4^(\circ)`

`F_1=20\ N`

`\tan (4)=(20)/(mg)----1`

for
`\theta =8^(\circ)`

Force is
`F_2`

`\tan (8)=(F_2)/(mg)---2`

Divide 1 and 2 we get

`(\tan (4))/(\tan (8))=(F_1)/(F_2)`

`F_2=20* (\tan 8)/(\tan 4)`

`F_2=20* 2.009`

`F_2=40.19\ N`