Static Equilibrium of hinge Question

  • Thread starter chara76
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  • #1
chara76
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Homework Statement



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.

http://img199.imageshack.us/img199/474/1235.gif" [Broken]

(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.


Homework Equations



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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Answers and Replies

  • #2
PhanthomJay
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You seem to be assuming that the force on the hinge acts in the direction of the sloped board axis. This is not correct. Use Newton 1 in the x and y directions to solve for the hinge force components.
 

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