Solution Found!
The 2.0-m-long, 15 kg beam in Figure P7.34 is hinged at
Chapter 7, Problem 22P(choose chapter or problem)
The \(2.0\)-m-long, \(15 \mathrm{~kg}\) beam in Figure P7.34 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. What is the gravitational torque on the beam about an axis through the hinged end when the beam is at the
a. Upper position?
b. Middle position?
c. Lower position?
Questions & Answers
QUESTION:
The \(2.0\)-m-long, \(15 \mathrm{~kg}\) beam in Figure P7.34 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. What is the gravitational torque on the beam about an axis through the hinged end when the beam is at the
a. Upper position?
b. Middle position?
c. Lower position?
ANSWER:
Step 1 of 3
(a) Upper Position:
The length of the beam: \(2 \mathrm{~m}\)
The distance from the centre of gravity to the axis: \(L=\frac{2}{2}=1 \mathrm{~m}\)
Perpendicular distance from the centre of gravity to the axis: \(r=1 \cos 20^{\circ}=0.93 \mathrm{~m}\) Gravitational torque on the beam through the axis:
\(\begin{aligned}\tau_{u}=r F \\\tau_{u}=-0.93 \times 15 \times 9.8=-136.71 \mathrm{Nm}
\end{aligned}\)
The gravitational torque on the beam about the axis in a clockwise direction is 136.71Nm.