**Tom and Jerry both push on the 3.10 m-diameter merry-go-round shown in the figure.**

**If Tom pushes with a force of 39.3 and Jerry pushes with a force of 35.5 , what is the net torque on the merry-go-round?**

**Known variables:**

**diameter: **3.10 m or

**radius: **1.55 m

**Force_1**= 39.3 N

**Force_2:** 35.5 N

Force_1 angle: 60**°**

Force_2 angle: 80**°**

First off, we would need to find torque of each person

the equation of torque is as follows: **τ** = Frsinθ

**τ**_{Tom} = 39.3 x 1.55 sin(60) = 52.75

**τ**_{Jerry} = 35.5 x 1.55 sin(80) = 54.19

Since we know that the direction of the rotation is clockwise, we can determine the signs for the torque.

**τ**_{Tom} = +52.75

**τ**_{Jerry} = -54.19

Just add the torques together to find the net torque.

**τ=** τ1 + τ2

**τ**= 52.75 + -54.19

**τ=** -1.44 N*m

**2) What is the net torque if Jerry reverses the direction that he pushes by 180without changing the magnitude of his force?**

If Jerry reverses the direction, then we would just have to add the torques.

**τ**= 52.75 + 54.19

**τ=**107 N*m

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