A 26kg cover for a roof opening is hinged at corners A and B

A 26-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 32 degree with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine the magnitude of the force exerted by the brace, the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust.

Solution

Force exerted by CE:

F = F*cos 74 deg i + F*sin 74 deg j

F = 0.2756*F i + 0.9612*F j

W = m*g = 26*9.81 = 255.06 N

r_A/B = 0.6 k

r_C/B = 0.9 i + 0.6 k

r_G/B = 0.45 i + 0.3 k

G = Centre of roof cover

F = F*(0.2756 i + 0.9612 j)

Total moment about hinge B:

Total Mb = 0

r_G/B x (-255.06 j) + r_C/B x F + r_A/B x A = 0

(0.45 i + 0.3 k) x (-255.06 j) = 0.45*(-255.06) k + 0.3*255.06 i = 76.518 i - 114.777 k

(0.9 i + 0.6 k) x (F*(0.2756 i + 0.9612 j)) = 0.9*0.9612*F k + 0.6*0.2756*F j - 0.6*0.9612*F i

(0.6 k) x (Ax i + Ax j) = 0.6*Ax j - 0.6*Ay i

Putting sum of every coefficient = 0

1. (76.518 - 0.6*0.9612*F - 0.6*Ay) i = 0

2. (0.6*Ax + 0.6*0.2756*F) j = 0

3. (0.9*0.9612*F - 114.777) k = 0

from 3

F = 114.777/(0.9*0.9612) = 132.678 N

from 2

Ax = -0.6*0.2756*132.678/0.6= -36.566 N

from 1

Ay = (76.518 - 0.6*0.9612*132.678)0.6 = 0 N

Now using force Balance:

Total F = 0 = A + B + F - W = 0

Ax + Ay + Bx + By + Bz + Fx + Fy - Wy = 0

Again Putting sum of every coefficient = 0

1. Ax + Bx + Fx = 0

2. Ay + By + Fy - Wy = 0

3. Bz = 0

Solving these equations:

Bz = 0

from 1

0.2756*132.678 - 36.566 + Bx = 0

0 + Bx = 0 N

Bx = 0 N

from 2

By + 0.9612*132.678 - 255.06 = 0

By = 255.06 - 0.9612*132.678 = 127.530 N

By = 127.530 N

Final Answer:

F = 132.678 N

Ay = -36.566 i N

B = 127.530 j N

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