Answer:
138 N-m
Explanation:
mg(cos 20°)(1.0) = 15(9.8)(cos 20°)(1.0) = 138 m-N
The gravitational torque on the beam is "[tex]-138.13 Nm.\hat{k}[/tex]"
According to the question,
We will have to take clockwise torque negative (-).
Now,
In the upper position,
→ [tex]\vec{\tau} = mg\times \frac{2}{2} Cos 20^{\circ}[/tex]
By substituting the values, we get
→ [tex]=15(9.8)(cos20^{\circ})(1.0)[/tex]
→ [tex]= 147\times 0.94[/tex]
→ [tex]= -138.13 \ Nm.\hat{k}[/tex]
Thus the above answer is appropriate.
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