The 2.0m -long, 15kg beam in the figure is hinged at its left end. It is \"falli
ID: 2184226 • Letter: T
Question
The 2.0m -long, 15kg beam in the figure is hinged at its left end. It is "falling" (rotating clockwise, under the influence of gravity), and the figure shows its position at three different times.(Figure 1.) What is the gravitational torque on the beam about an axis through the hinged end when the beam is at the upper position? Express your answer using two significant figures. What is the gravitational torque on the beam about an axis through the hinged end when the beam is at the middle position? Express your answer using two significant figures. What is the gravitational torque on the beam about an axis through the hinged end when the beam is at the lower position? Express your answer using two significant figures.Explanation / Answer
In the middle (horizontal) position, the torque is (15 x 1) = 15N/m. This is because the CG is 1m. from the pivot (in the beam centre). In the upper position at 20 deg., you need the horizontal distance the beam extends. (cos 20) x 2m = 1.8794m. Now, halve that, and the CG is 0.9397m. horizontally, from the pivot. Torque = (15 x .9397) = 14.0955N/m
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