Problem 5. A rigid beam of mass m, length L, has two supports. One support is lo

Question

Problem 5. A rigid beam of mass m, length L, has two supports. One support is located at the beam's left end, while the other is a distance 3L/4 from the left edge. A block of mass M is situated such that its center of mass is located a distance L/2 from the beam's left end as showin in the figure below. What is the ratio of the normal force exerted by support 1 to the normal force exerted by support 2 on the beam? Block of mass M Rigid Beam of mass m, length L L/2 Support 1 Support 2 3L/4 F2 4 C) F2 4 D) E) F2

Explanation / Answer

D) F1/F2 = 1/2

As the beam is in equilibrium net force and torque acting on it must be zero.

Apply net torque about left end = 0

m*g*(L/2)*sin(90) + M*g*(L/2)*sin(90) - F2*(3*L/4)*sin(90) = 0

m*g/2 + M*g/2 - F2*3/4 = 0

m*g + M*g - F2*3/2 = 0

==> F2*3/2 = (m + M)*g

F2 = (2/3)*(m + M)*g ---(1)

now apply Fnety = 0

F1 + F2 - m*g - M*g = 0

F1 = (m + M)*g - F2

= (m + M)*g - (2/3)*(m + M)*g

= (1/3)*(m + M)*g ---(2)

from equations 1 and 2

we get,

F1/F2 = (1/3)/(2/3)

F1/F2 = 1/2

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.