Consider a rigid 2.70-m-long beam (see figure) that is supported by a fixed 1.35

Question

Consider a rigid 2.70-m-long beam (see figure) that is supported by a fixed 1.35-m-high post through its center and pivots on a frictionless bearing at its center atop the vertical 1.35-m-high post. One end of the beam is connected to the floor by a spring that has a force constant k = 1300 N/m. When the beam is horizontal, the spring is vertical and unstressed. If an object is hung from the opposite end of the beam, the beam settles into an equilibrium position where it makes an angle of 22.4° with the horizontal. What is the mass of the object?

__________________kg

Explanation / Answer

if the rod system remains to be in equilibrium

it must happen that sum of all forces and torque will be zero

so if the force applied by the spring is F

then ets Calculate the Torque about pivot point as

F d cos theta = r x F

so

mg*0.5 L cos 22.4 = r*F sin theta

for Sprng , Force F = kx

so

4.53 mL = 0.5 L* k x sin theta

4.53 mL =0.5 * 1300 * L * Sin theta

4.52 mL = 650 L*sin 22.4

m = 247.69/(4.52)

m = 54.79 kgs

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