A stationary 3.0-m board of mass 6.0 kg is hinged at one end. A force is applied vertically at the other end, and the board makes a 30° angle with the horizontal. A 50-kg block rests on the board 80 cm from the hinge as shown in the figure below.
(a) Find the magnitude of the force . Answer is F=160.23 N
(b) Find the force exerted by the hinge. Fh=389.13 N
(c) Find the magnitude of the force as well as the force exerted by the hinge, if the magnitude of the force is exerted, instead, at right angles to the board.
The answer to the first part is 138.763 N. The second part is the one I cannot figure out.
Fx: Force of hinge*cos(theta) = F*sin(theta)
Fy: Force of hinge*sin(theta) = (M+m)g - F*cos(theta)
The Attempt at a Solution
Fh*sin(theta)/Fh*cos(theta) = [(6+50)(9.81)-(160.23)*cos(3)] / 160.23*sin(30) = 5.12509
theta = arctan(5.12509) = 78.959 degrees
Fh = [(160.23)*sin(30)]/cos(78.959) = 418.33 N (However, this answer is wrong)
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