Question

9 gary and clark are pulling a crate using ropes. gary pulls with a force of 110 n, 25° west of north. clark pulls with a force of 95 n, 10° east of north. what is the net force gary and clark are exerting on the crate? 174 n, 12° east of north 201 n, 14° west of north 159 n, 7.0° east of north 196 n, 8.8° west of north

Explanation:

Step1: Resolve forces into components

For Gary's force \(F_G = 110\space N\), \(\theta_G=25^{\circ}\) west of north.
North - component: \(F_{G,y}=F_G\cos(25^{\circ})=110\cos(25^{\circ})\approx99.77\space N\)
West - component: \(F_{G,x}=-F_G\sin(25^{\circ})\approx - 110\times0.4226=-46.49\space N\)

For Clark's force \(F_C = 95\space N\), \(\theta_C = 10^{\circ}\) east of north.
North - component: \(F_{C,y}=F_C\cos(10^{\circ})\approx95\times0.9848 = 93.56\space N\)
East - component: \(F_{C,x}=F_C\sin(10^{\circ})\approx95\times0.1736 = 16.59\space N\)

Step2: Sum the components

Total north - component: \(F_y=F_{G,y}+F_{C,y}=99.77 + 93.56=193.33\space N\)
Total east - component: \(F_x=F_{C,x}+F_{G,x}=16.59-46.49=-29.9\space N\) (negative \(x\) - component means net west direction)

Step3: Calculate the magnitude of the net force

\(F=\sqrt{F_x^{2}+F_y^{2}}=\sqrt{(- 29.9)^{2}+193.33^{2}}=\sqrt{894.01 + 37376.49}=\sqrt{38270.5}\approx196\space N\)

Step4: Calculate the direction

\(\tan\theta=\frac{\vert F_x\vert}{F_y}=\frac{29.9}{193.33}\)
\(\theta=\arctan(\frac{29.9}{193.33})\approx8.8^{\circ}\) west of north

Answer:

196 N, \(8.8^{\circ}\) west of north