# Chapter 12 - Exercises and Problems - Page 215: 28

(a) $1.96KN$ (b) $2.6KN$

#### Work Step by Step

We know that the net torque is zero $\implies Tcos25^{\circ}\times 4.55-N\times 11.6=0$ This simplifies to: $T=\frac{N\times 11.6}{4.55cos25^{\circ}}=\frac{697\times 11.6}{4.55cos25^{\circ}}=1.96KN$ (b) In the given case $\sum F_x=0$ $\implies Tsin25^{\circ}=F_{cx}$ $\implies F_{cx}=1.96\times sin25=829N$ and $\sum F_y=0$ $\implies Tcos25^{\circ}+N=F_{cy}$ $F_{cy}=1.96\times cos25^{\circ}+697$ $F_{cy}=1.96cos25^{\circ}+697=2.5KN$ Now $F_c=\sqrt{F_{cx}^2+F_{cy}^2}$ We plug in the known values to obtain: $F_c=\sqrt{(829)^2+(2.5)^2}=2.6KN$

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