Consider a rigid 2.90-m-long beam (see figure) that is supported by a fixed 1.45

Question

Consider a rigid 2.90-m-long beam (see figure) that is supported by a fixed 1.45-m-high post through its center and pivots on a frictionless bearing at its center atop the vertical 1.45-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 24.9° with the horizontal. What is the mass of the object? 84 Draw an extended free-body diagram of the forces acting on the beam. Because the beam is in rotational equilibrium, the sum of the torques acting on it must equal zero. Be careful with your geometry when determining the extended length of the spring. You will need to use the law of cosines and some trigonometric identities

Explanation / Answer

let mass of the object be m
now, theta = 24.9 deg
let length of the beam be 2a = 2.9 m
also, L = 1.45 m
spring constant, k = 1500 N/m
new length of the spring = l
l = sqroot((L + asin(theta))^2 + (a - acos(theta))^2)
l = sqroot((1.45 + 1.45sin(24.9))^2 + (1.45 - 1.45cos(24.9))^2)
l = 2.0649 m
hence tension in the spring = k(l - L) = 922.3585 N
now, Tcos(phi) = mg ( from torque balance)
cos(phi) = (L + asin(theta))/l = (1.45 + 1.45*sin(24.9))/2.0649
phi = 3.7402 deg
hence
m = Tcos(phi)/g = 93.822 kg

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---