Question

3 -1 POINTS OSUNIPHYS1 12.2.WA.018. MY NOTES | ASK YOUR TEACHER As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of 30°, the length of the beam is L-1.75 m, the coefficient of static friction between the wall and the beam is 0.540, and the weight of the beam is represented by w. Determine the minimum distance x from point A at which an additional weight w twice the weight of the rod) can be hung without causing the rod to slip at point A. Additional Materials

Answer 1

Horizontally, for tension T and normal force on beam F_n

F_n = Tcos30º = 0.866T

the threshold, F_f = µ x F_n = 0.540 x 0.866T = 0.468T

where F_f: the friction force at the beam

Vertically, we've got

F_f + Tsin30º = 0.468T + 0.5T = 0.968T

0.968T = w + 2w = 3w

so T = 3.099w

Finally, consider the moment about the left end of the beam.

It must be zero, or the beam would rotate.

ΣM = 0 = Tsin30º ( L )- w(L/2) - 2wx

=> (3.099w )( 0.5 )( L) - w(L/2) - 2wx = 0

=>(1.5495L - L/2 - 2x)w= 0

=>1.5495L - L/2 - x = 0

=>1.0495L - 2x = 0

=>2x = 1.0495L

=>x = (1.0495/2)L = 0.52475L

Since L = 1.75 m

So, x = 0.52475 x 1.75 = 0.918 m

Hope this will help you. Please comment if there is any doubt.

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