Question 3.20: Determine the forces exerted on the cylinder at A and B by t...

As the applied force is acting on spanner wrench i.e., 360 kN, thus considering free body diagram of the same as shown in Fig. 3.31 (a)
Using equations,

ΣX = 0,

R_{A H}=R_{B}                                      ….. (1)

ΣY = 0,

R_{A V} = 360 KN

\sum M_{A}=0

R_{B} \times 0.5=360 \times 3

R_{B} = 2160 KN      Thus from equation(1),

R_{A H} = 2160  KN

\therefore \quad R_{A}=\sqrt{R_{A H}^{2}+R_{A V}^{2}}=\sqrt{(2160)^{2}+(360)^{2}}

R_{A } = 2189.79 KN

Thus these reactions will be applied on cylinder but in opposite directions as shown below

R_{B} = 2160 KN

R_{A } = 2189.79 KN

\tan \alpha=\frac{R_{A V}}{R_{A H}}

=\frac{{360}}{{ 2160}}

\alpha = 9.46°              in fourth quadrant as shown above.