A 5 m board of mass 8 kg is hinged at one end, with a 56 kg block nesting on the
ID: 2303361 • Letter: A
Question
A 5 m board of mass 8 kg is hinged at one end, with a 56 kg block nesting on the board 193 cm from the hings. A force F is applied vertically at the other end to lift of the board.
A) Find the magnitude of the force needed to hold the board stationary at an angle of elevation of 35 degrees. Assume the block is a point particle. Gravity is 9.81
B) Find the force exerted by the hinge at this angle
C) Find the magnitude of the force F if F is exerted perpendicular to the board when the angel of elevation of the board is 35 degrees
D) Find the force exerted by the hinge if F is exerted perpendicular to the board when the angle of elevation of the board is 35 degrees
Explanation / Answer
a) along vertical
m1*g + m2*g + Fy + F = 0
along horizontal
Fx = 0
net torque = 0
m1*g*L/2*cos34 + m2*g*r*cos34 = F*L*cos34
(8*9.8*2.5*cos35) + (56*9.8*1.93*cos35) = F*5*cos35
F = 251.037 N
b) along horizontal
Fx = 0
Fy + F - m1g - m2g = 0
Fy = 376.163 N
c)
net torque = 0
m1*g*L/2*cos35 + m2*g*r*cos35 = F*L*sin90
(8*9.8*2.5*cos35) + (56*9.8*1.93*cos35) = F*5
F = 205.637N
d) Fx - F*cos(90-35) = 0
Fx = 117.949 N
Fy + F*sin(90-35) - m1*g - m2*g = 0
Fy = 458.752 N
F at hinge = sqrt(Fx^2 +Fy^2) = 473.672 N
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