Consider a rigid 2.10-m-long beam (see figure) that is supported by a fixed 1.05
ID: 1469577 • Letter: C
Question
Consider a rigid 2.10-m-long beam (see figure) that is supported by a fixed 1.05-m-high post through its center and pivots on a frictionless bearing at its center atop the vertical 1.05-m-high post. One end of the beam is connected to the floor by a spring that has a force constant k = 1500 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 20.2° with the horizontal. What is the mass of the object?
kg
Explanation / Answer
Take summation of moments about the center (where the beam pivots on a frictionless bearing).
mg(cos 20.2)(L/2) = kx(L/2)
where
m = mass
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
L = length of the beam
k = spring constant
x = length of stretched portion of spring
Solving for "m",
m = kx/(g*cos 20.2) --- call this Equation 1
From the geometry of the figure,
tan 20.2 = x/(L/2) = x/1.05
Solving for x,
x = 1.05(tan 20.2) = 0.38 m.
Substituting the value of "x" in Equation 1,
m = (1500)(0.38)/(9.8 * cos 20.2)
m = 61.97 kg. = 62 kg