Question

It is desired to remove the spike from the timber by applying force along its horizontal axis. An obstruction \(A\) prevents direct access, so that two forces, one \(370 \mathrm{lb}\) and the other \(\mathbf{P}\), are applied by cables as shown. Compute the magnitude of \(\mathbf{P}\) necessary to ensure a resultant \(\mathbf{T}\) directed along the spike. Also find \(T\).

Answer 1

Angle which the force P makes with center of the spike( shown by dotted line)

\(\theta=\tan ^{-1}\left(\frac{5^{\prime \prime}}{10^{\prime \prime}}\right)=26.565^{\circ}\)

Similarly Angle made by force \(370 \mathrm{lb}\) with the dotted line( below that line) is

\(\phi=\tan ^{-1}\left(\frac{8^{\prime \prime}}{10^{\prime \prime}}\right)=38.66^{\circ}\)

Then since there has to no motion in vertical direction then balancing force in vertical direction,

\(P \sin \theta=F \sin \phi\)

using given and calculated values in above,

\(P \sin 26.56^{\circ}=(370 l b) \sin 38.66^{\circ}\)

\(P=516.931 l b\) (ANS)

Now balancing forces in horizontal direction,

\(T=P \cos \theta+F \cos \phi\) (Since "T" is directed towards spike or towards left)

using calculated and given values in above,

\(T=(516.931) \cos 26.56^{\circ}+(370) \cos 38.66^{\circ}\)

\(T=462.377+288.92=751.297 l b\) (ANS)