A stationary 3.0-m board of mass 4.5 kg is hinged at one end. A force F with arr

Question

A stationary 3.0-m board of mass 4.5 kg is hinged at one end. A force F with arrow 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 F with arrow. 152.716 Correct: Your answer is correct. N (b) Find the force exerted by the hinge. 1154.29 Incorrect: Your answer is incorrect. N (c) Find the magnitude of the force F with arrow as well as the force exerted by the hinge (F with arrowH), if F with arrow is exerted, instead, at right angles to the board. F with arrow 132.25 Correct: Your answer is correct. N F with arrowH 534.1 Incorrect: Your answer is incorrect. N

Explanation / Answer

<< Find the magnitude of the force vector F >>

Take summation of moments by the hinge,

F(cos 30)(3) = 4.5(9.8)(cos 30)(1.5) + 50(9.8)(0.8)(cos 30)

From the above, I trust that you can solve for "F"

.
<< Find the force exerted by the hinge. >>

Let

Fv = vertical force exerted by the hinge
Fh = horizontal force exerted by the hinge

Fv = F + (50 + 4.5)(9.8)

Fv = F + 54.5(9.8)

where

F = force determined above

and calculate for the corresponding values of Fv.

Summation of horizontal forces,

Fh = 50(9.8)(sin 30) + 4.5(9.8)(sin 30) + F(sin 30)

From the above, calculate the value of Fh.


<< Find the magnitude of the force vector F as well as the force exerted by the hinge (vector F H), if vector F is exerted, instead, at right angles to the board >>

Following the above procedure.

Summation of moment about the hinge,

F(3) = 4.5(9.8)(cos 30)(1.5) + 50(9.8)(0.8)(cos 30)

and solve for F. =396.769 N

For the vertical and horizontal components of the reactions in the hinge, simply take the summation of the vertical and horizontal components of the forces (as in the above procedure) acting on the system

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