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Nesterboy [21]
3 years ago
9

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

an axis of rotation perpendular to the square and through the center. The axis of rotation passes through the center and perpendicular to the square.

Physics
1 answer:
marshall27 [118]3 years ago
3 0

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

\tau_{1}=-18\times\dfrac{0.2}{2}

\tau_{1}=-1.8\ N-m

We need to calculate the torque due to force F₂

Using formula of torque

\tau_{2}=F_{2}d_{2}

\tau_{2}=F_{2}\times\dfrac{a}{2}

Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

\tau_{2}=2.6\ N-m

We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}

Put the value into the formula

\tau_{3}=0.1(14\sin45+14\cos45)

\tau_{3}=1.92\ N-m

We need to calculate the net torque on the square plate

\tau=\tau_{1}+\tau_{2}+\tau_{3}

\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

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