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Question

Plot the magnitude of the resultant R of the three forces as a function of θ\theta for 0θ3600 \leq \theta \leq 360^{\circ} and determine the value of θ\theta which makes the magnitude R of the resultant of the three loads (a) a maximum and (b) a minimum. Record the magnitude of the resultant in each case. Use values of ϕ=75 and ψ=20\phi=75^{\circ} \text { and } \psi=20^{\circ}.

Solution

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First we will write the forces as vectors.

F1=85(cosϕsinθi+cosϕcosθj+sinϕk) N=85(cos75°sinθi+cos75°cosθj+sin75°k) N=(22sinθi+22cosθj+82.10k) NF2=105(0icosψj+sinψk) N=105(0icos20°j+sin20°k) N=(0i98.67j+35.91k) NF3=60(1i+0j+0k) N=(60i+0j+0k) N\begin{align*} \vec{\mathbf{F}_1}&=85(-\cos\phi\sin\theta\vec{\mathbf{i}}+\cos\phi\cos\theta\vec{\mathbf{j}}+\sin\phi\vec{\mathbf{k}})\text{ N}\\ &=85(-\cos75\text{\textdegree}\sin\theta\vec{\mathbf{i}}+\cos75\text{\textdegree}\cos\theta\vec{\mathbf{j}}+\sin75\text{\textdegree}\vec{\mathbf{k}})\text{ N}\\ &=(-22\sin\theta\vec{\mathbf{i}}+22\cos\theta\vec{\mathbf{j}}+82.10\vec{\mathbf{k}})\text{ N}\\ \vec{\mathbf{F}_2}&=105(0\vec{\mathbf{i}}-\cos\psi\vec{\mathbf{j}}+\sin\psi\vec{\mathbf{k}})\text{ N}\\ &=105(0\vec{\mathbf{i}}-\cos20\text{\textdegree}\vec{\mathbf{j}}+\sin20\text{\textdegree}\vec{\mathbf{k}})\text{ N}\\ &=(0\vec{\mathbf{i}}-98.67\vec{\mathbf{j}}+35.91\vec{\mathbf{k}})\text{ N}\\ \vec{\mathbf{F}_3}&=60(1\vec{\mathbf{i}}+0\vec{\mathbf{j}}+0\vec{\mathbf{k}})\text{ N}\\ &=(60\vec{\mathbf{i}}+0\vec{\mathbf{j}}+0\vec{\mathbf{k}})\text{ N}\\ \end{align*}

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