Math

QuestionA uniform beam of length, LL and mass, MM, is freely pivoted at one end about an attachment point in a wall. The other end is supported by a horizontal cable also attached to the wall so that the beam makes an angle ϕ\phi with the horizontal as shown below.
If L=0.9 m.M=15 kgL=0.9 \mathrm{~m} . M=15 \mathrm{~kg}. and ϕ=30\phi=30^{\circ}, then what is the magnitude of the torque due to gravity about the pivot point? \square Nm
If the TT \square radss2\mathrm{rads} \mathrm{s}^{-2} HINT: The moment of intertia for a uniform beam about one end is given by I=13ML2I=\frac{1}{3} M L^{2}.

Studdy Solution
The magnitude of the torque due to gravity about the pivot point is 33.075 Nm33.075 \text{ Nm}.

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