1. A force P is applied to one end of a beam with a length of 2 m. What is the magnitude of the torque? The axis of rotation at point A.

__Known :__

Force (F) = 10 N

Length of AB (r_{AB}) = 2 m

*Force F is perpendicular to the beam. *

The lever arm (l) = r_{AB }sin 90^{o }= (2 m)(1) = 2 m

__Wanted:__ The torque about the axis of rotation

__Solution :__

The torque :

τ = F l = (10 N)(2 m) = 20 N m

*The plus sign because the beam **rotates counterclockwise rotation.** *

2. The length of a beam AB is 2 m and the magnitude of force F is 10 N. What is the magnitude of the torque? The axis of rotation at point A.

__Known :__

Force (F) = 10 N

Length of AB (r_{AB}) = 2 m

The lever arm (l) = r_{AB }sin 60^{o }= (2 m)(0.5√3) = √3 m

__Wanted:__ The torque about the axis of rotation

__Solution :__

The torque :

τ = F l = (10 N)(√3 m) = 10√3 N m

*The plus sign because the force F causes the beam **rotates counterclockwise rotation.** *

3. The length of a beam is 2 m. The magnitude of F_{1} is 10 N and the magnitude of F_{2} is 15 N. Determine the net torque about the center of the beam.

*The axis of rotation at the center of the beam.*

__Known :__

Force 1 (F_{1}) = 10 N

The distance between F_{1} and the center of beam (r_{1}) = 1 m

The lever arm 1 (l_{1}) = r_{1 }sin 90^{o }= (1 m)(1) = 1 m

*Force F*_{1}* is perpendicular to the beam. *

Force 2 (F_{2}) = 15 N

The distance between F_{2} and the center of the beam (r_{2}) = 1 m

*Force F*_{2}* is perpendicular to the beam. *

The lever arm 2 (l_{2}) = r_{1 }sin 90^{o }= (1 m)(1) = 1 m

__Wanted :__ The net torque about the axis of rotation

__Solution :__

The torque 1 :

τ_{1} = F_{1} l_{1} = (10 N)(1 m) = 10 N m

*The plus sign because the force of F*_{1 }*causes the beam **rotates counterclockwise rotation.** *

The torque 2 :

τ_{2} = F_{2} l_{2 }= (15 N)(1 m) = -15 N

*The minus sign because the force F*_{2}* causes the beam to **rotates clockwise.** *

The net torque :

Στ = τ_{1 }– τ_{2 }= 10 – 15 = – 5 N m

*The minus sign because the beam to **rotates clockwise.** *

4. The length beam AB is 2 m, The magnitude of F_{1} is 10 N and the magnitude of F_{2 }is 10 N. Determine the net torque about the center of the beam.

*The axis of rotation at the center of the beam.*

__Known :__

Force 1 (F_{1}) = 10 Nn

The distance between F_{1} and the center of the beam (r_{1}) = 1 m

The lever arm 1 (l_{1}) = r_{1 }sin 60^{o }= (1 m)(0.5√3) = 0.5√3 m

Force 2 (F_{2}) = 10 N

The distance between F_{2} and the center of the beam (r_{2}) = 1 m

*Force F*_{2}* is perpendicular to the beam. *

The lever arm 2 (l_{2}) = r_{2 }sin 90^{o }= (1 m)(1) = 1 m

__Wanted :__ The net torque about the axis of rotation

__Solution :__

The torque 1 :

τ_{1} = F_{1} l_{1} = (10 N)(0.5√3 m) = 5√3 = 8.7 N.m

*The plus sign because the force F*_{1}* causes the beam to **rotates clockwise.** *

The torque 2 :

τ_{2} = F_{2} l_{2 }= (10 N)(1 m) = -10 N m

*The minus sign because the force F*_{2}* causes the beam to **rotates clockwise.*

The net torque :

Στ = τ_{1 }– τ_{2 }= 8.7 – 10 = – 1.3 N m

*The minus sign because the net force causes the beam to **rotates clockwise.** *

5. The length of a beam is 10 m, the magnitude of F_{1 }is 10 N, the magnitude of F_{2} is 10 N and the magnitude of F_{3} is 15 N. The distance between point A and point C is 7.5 m. The Force F_{2} located at the center of the beam. Determine the net torque about the point C located at 2.5 m from the point B.

*The axis of rotation located at point C*

__Known :__

Force 1 (F_{1}) = 10 N

The distance between F_{1} and point C (r_{1}) = 2.5 m

*Force F*_{1}* is perpendicular to the beam. *

The lever arm 1 (l_{1}) = r_{1 }sin 90^{o }= (2.5 m)(1) = 2.5 m

Force 2 (F_{2}) = 10 N

The distance between F_{2} and point C (r_{2}) = 2.5 m

*The force F*_{2}* is perpendicular to the beam. *

The lever arm 2 (l_{2}) = r_{2 }sin 90^{o }= (2.5 m)(1) = 2.5 m

Force 3 (F_{3}) = 15 N

The distance between F_{3} and point C (r_{3}) = 7.5 m

*The force F*_{3}* is perpendicular to the beam. *

The lever arm 3 (l_{3}) = r_{3} sin 90^{o }= (7.5 m)(1) = 7.5 m

__Wanted :__ The net torque about the axis of rotation

__Solution :__

The torque 1 :

τ_{1} = F_{1} l_{1} = (10 N)(2.5 m) = 25 N m

*The plus sign because the force F*_{1}* causes the beam to **rotates clockwise.** *

The torque 2 :

τ_{2} = F_{2} l_{2 }= (10 N)(2.5 m) = 25 N m

T*he plus sign because the force F*_{2}* causes the beam to **rotates clockwise.** *

The torque 3 :

τ_{3} = F_{3} l_{3} = (15 N)(7.5 m) = -112..5 N m

*The minus sign because the force F*_{3}* causes the beam to **rotates clockwise.** *

The net torque :

Στ = τ_{1 }+ τ_{2 }– τ_{3 }= 25 + 25 – 112.5 = – 62.5 N m

*The minus sign because the net force causes the beam to **rotates clockwise.** *

6. The length of a beam is 10 m, the magnitude of F_{1} is 10 N, the magnitude of F_{2} is 10 N and the magnitude of F_{3} is 10 N. Determine the net torque about point A, located 5 m from the point of application of force F_{1}.

*The axis of rotation at point A.*

__Known :__

Force 1 (F_{1}) = 10 N

The distance between F_{1} and point A (r_{1}) = 5 m

The lever arm 1 (l_{1}) = r_{1 }sin 60^{o }= (5 m)(0.5√3) = 2.5√3 m

Force 2 (F_{2}) = 10 N

The distance between F_{2} and point A (r_{2}) = 0

*The force F*_{2}* is perpendicular to the beam.*

The lever arm 2 (l_{2}) = r_{2 }sin 90^{o }= (0)(1) = 0

The force 3 (F_{3}) = 10 N

The distance between F_{3} and point A (r_{3}) = 10 m

The lever arm 3 (l_{3}) = r_{3 }sin 30^{o }= (10 m)(0.5) = 5 m

__Wanted :__ the net torque about the axis of rotation

__Solution :__

The torque 1 :

τ_{1} = F_{1} l_{1} = (10 N)(2.5√3 m) = 25√3 = 43.3 N m

*The plus sign because the force F*_{1}* causes the beam to **rotates clockwise.** *

The torque 2 :

τ_{2} = F_{2} l_{2 }= (10 N)(0) = 0

The torque 3 :

τ_{3} = F_{3} l_{3} = (10 N)(5 m) = -50 N m

*The minus sign because the force F*_{3}* causes the beam to **rotates clockwise.** *

The net torque :

Στ = τ_{1 }+ τ_{2 }– τ_{3 }= 43.3 + 0 – 50 = – 6.7 N m

The minus sign because the net force causes the beam to rotates clockwise.