Physics - Equilibrium: How would I do this question? A mass m hangs by a cable that is attached to an L shaped rigid bar by a pin at point C. The cable passes over a frictionless pulley as shown. The bar is massless and loaded and supported as shown. Determine the reaction force (magnitude and direction) at A and the mass m such that the bar is in static equilibrium. Image:

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For static equilibrium the sum of force moments with respect to point A, and the sum of forces on the horizontal and vertical axis need to be zero.

Sum of moments with respect to point A:

`F_c*cos(30)*12 +10*2 -100 -5*5 =0`

`F_c =(100+25-20)/(12*cos(30)) =10.1 N`

Sum of forces on x axis:

`F_C*sin(30) +5 -F_(xA) =0`

`F_(xA) =5+10.1*sin(30) =10.05 N`

Sum of forces on y axis:

`F_c*cos(30) +10-F_(yA) =0`

`F_(yA) =10+10.1*cos(30) =18.75 N`

`|F_A| =|10.05*hati +18.75*hatj| =sqrt(10.05^2 +18.75^2) =21.27 N`
To determine the value of the mass m, see the insert figure in the upper left corner.

`G` decomposes on to horizontal and vertical components along the pulley direction (45 degree):

`G_p =G*sin(45)`   and  `G_n=G*cos(alpha)`

The component `G_n` is canceled by the pulley normal reaction. The component `G_p` further decomposes along the wire at 60 degrees and perpendicular to the wire at 60 degrees (this last component is canceled by the fact that the pulley does not move along the inclined plane).

`F_C =G_p*cos(60-45) =m*g*sin(45)*cos(15) `

Therefore the value of the mass is

`m = F_C/(g*sin(45)*cos(15)) =10.1/(9.81*sin(45)*cos(15)) =1.507 kg`

Answer: the reaction at A is 21.27 N and the mass m is 1.5 kg.

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