Problem based on Rigid Body Mechanics (Newtonian-Mechanics)

Question

I got this problem in a tutorial sheet given to me by my Professor. I am confused as to how to apply the torque about point C, which is hinged. Please be kind to ignore any wrong format of asking the question. Without any further ado, Here's the question:

The rigid triangular element shown is hinged at C and supported at pin B (BC is horizontal). Assuming all contacts to be frictionless, given F=84.2 N in the direction shown, calculate

(a) the magnitude of reaction force at pin B, and

(b) the magnitude of the net reaction force at hinge C.

Answer 1

$F_X=842 cos(15)^o=813, N$

$F_y=842 sin (15^o)=218, N$

$T_x=813 \times 75=61000 N-mm=61 N-m$

$T_y=218\times (75+150)=49033 N-mm=49 N-m$

Total moment(torque) on C:

$T_C=61+49=110 N-m$

This torque is resisted by reaction force on B, so total reaction force on B is:

$F_B=\frac{110}{0.15}=734 , N $

This force is $y^-$ direction and can be considered as the vector sum of two components because of slot.

The external force to system is $842, N$ which can be considered as total force on C.

Answer 2

Calling

$$ \cases{ A = (x_a,y_a)\\ B = (x_b, y_b)\\ C = (x_c, y_c)\\ F_A = f_a(-\cos(\alpha_1),\sin(\alpha_1))\\ F_B = f_b(\cos(\alpha_2),\sin(\alpha_2))\\ F_C = (f_{c_x},f_{c_y}) } $$

with

$$ \cases{ \alpha_1 = \frac{\pi}{12}\\ \alpha_2 = \frac{\pi}{6}\\ f_a = 84.2N\\ x_a =-0.075m\\ y_a =-0.075m\\ x_b =0m\\ y_b=0m\\ x_c=0.150m\\ y_c=0m } $$

The equilibrium equations are

$$ \cases{ F_A+F_B+F_C = 0\\ F_A\times(A-B)+F_C\times (C-B) = 0 } $$

or equivalently

$$ \left\{ \begin{array}{l} f_a \sin (\alpha_1) (x_a-x_c)+f_a \cos (\alpha_1) (y_a-y_c)+f_b \sin (\alpha_2) (x_b-x_c)+f_b \cos (\alpha_2) (y_c-y_b)=0 \\ f_b \cos (\alpha_2)+2 f_{c_x}=0 \\ f_b \sin (\alpha_2)+2 f_{c_y}=0 \\ \end{array} \right. $$

with solution

$$ \left\{ \begin{array}{l} f_b= \frac{f_a (\sin (\alpha_1) (x_c-x_a)+\cos (\alpha_1) (y_c-y_a))}{\sin (\alpha_2) (x_b-x_c)+\cos (\alpha_2)(y_c-y_b)} = -146.71N\\ f_{c_x}= \frac{f_a \cos (\alpha_2) (\sin (\alpha_1) (x_a-x_c)+\cos (\alpha_1) (y_a-y_c))}{2 \sin (\alpha_2)(x_b-x_c)+2 \cos (\alpha_2) (y_c-y_b)}= 63.53N \\ f_{c_y} =\frac{f_a \sin (\alpha_1) (x_a-x_c)+f_a \cos (\alpha_1) (y_a-y_c)}{2 (\cot (\alpha_2)(y_c-y_b)+x_b-x_c)}=36.68N \\ \end{array} \right. $$

or

$$ |f_b| = 146.71N,\ \ |f_c| = 73.35N $$

NOTE

$F_C$ is normal to the groove surface.