a uniform beam of length L=3.3m and mass M=42kg has its lower end fixed to a piv
ID: 1470789 • Letter: A
Question
a uniform beam of length L=3.3m and mass M=42kg has its lower end fixed to a pivot at a point P on the floor, making an angle of 28 degrees. A horizontal cable is attached at its upper end B to a point A on a wall. A box of the same mass M as the beam is suspended from a rope that is attached to the beam 1/4 L from its upper end.
a. What is the y-component Py of the force in newtons exerted by the pivot on the beam?
b. Write an expression for the tension T in the horizontal cable AB.
c. What is the x-component Px of the force in newtons exerted on the beam?
Explanation / Answer
a)
along vertical net force = 0
Py - Mg - Mg = 0
Py = 2*M*g = 2*42*9.8 = 823.2 N
b)
net torque about the pivot = 0
T*L*sin28 = M*g*L/2*cos28 + M*g*3L/4*cos28
T*sin28 = (42*9.8*0.58cos28)+(42*9.8*(3/4)*cos28)
T = 1029.56 N
c)
along horizantla net force = 0
Px - T = 0
Px = 1029.56 N
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