College Physics : Torque

Study concepts, example questions & explanations for College Physics

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Example Questions

Example Question #11 : Forces

A mechanic is able to exert a maximum torque of \(\displaystyle 144 inch\cdot lbs\) using a small wrench. The mechanic adds an attachment to the wrench that doubles the handle length. What is the maximum amount of torque the mechanic can exert with the new wrench?

Possible Answers:

\(\displaystyle 10736inch\cdot lbs\)

\(\displaystyle 144inch\cdot lbs\)

\(\displaystyle 72inch\cdot lbs\)

\(\displaystyle 288 inch\cdot lbs\)

\(\displaystyle 12inch\cdot lbs\)

Correct answer:

\(\displaystyle 288 inch\cdot lbs\)

Explanation:

The maximum amount of torque is the magnitude of the torque vector \(\displaystyle |\vec{r}\times\vec{F}|\) , which is maximized when the angle between the applied force and the displacement vector is 90 degrees. Then the torque is given by \(\displaystyle rF\), where \(\displaystyle r\) is the length of the wrench and \(\displaystyle F\) is the magnitude of the force applied. Since the mechanic doesn't get any stronger, but the handle length increases, the torque doubles.

Example Question #51 : Mechanics

An object weighing \(\displaystyle 20\:kg\) sits on a board at a distance \(\displaystyle 5\:m\) away from a fulcrum. At what distance to the right of the fulcrum would an \(\displaystyle 18\:kg\) object need to be in order to make the board flat?

Possible Answers:

\(\displaystyle 3.14\:m\)

\(\displaystyle 4.44\:m\)

\(\displaystyle 5.56\:m\)

\(\displaystyle 6.82\:m\)

Correct answer:

\(\displaystyle 5.56\:m\)

Explanation:

For this question, we'll need to take torque into account. The amount of torque caused by the object to the left of the fulcrum needs to be equal to the torque of the object to the right of the fulcrum.

\(\displaystyle \tau_{Left} =\tau_{Right}\)

Moreover, we can write out the equation for torque.

\(\displaystyle \tau =Fdsin(\theta)\)

\(\displaystyle F=mg\)

Combining these equations, we can set an expression for the torque on the left and the torque on the right.

\(\displaystyle \tau_{L} =m_{L}gd_{L}sin(\theta)\)

\(\displaystyle \tau_{R} =m_{R}gd_{R}sin(\theta)\)

Then we can set them equal to each other.

\(\displaystyle m_{L}gd_{L}sin(\theta)=m_{R}gd_{R}sin(\theta)\)

Next, we isolate the term for the distance on the right and we cancel out common units.

\(\displaystyle d_{R}=\frac{m_{L}gd_{L}sin(\theta)}{m_{R}gsin(\theta)}=\frac{m_{L}d_{L}sin(\theta)}{m_{R}sin(\theta)}\)

Finally, we plug in the values given in the question stem. Because we're told that both objects are sitting flat on the board, we know the force of their weight is perpendicular to the board, which gives a value of \(\displaystyle 90^{o}\).

\(\displaystyle d_{R}=\frac{(20\:kg)(5\:m)sin(90^{o})}{(18\:kg)sin(90^{o})}\)

\(\displaystyle d_{R}=5.56\:m\)

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