Question 5.3: The picture shows three men involved in moving a packing cas...

i) The diagram shows all the forces acting on the packing case and the relevant angles.

ii) Equilibrium equations

Resolving horizontally ( → )

T_{1}  cos  β  +  T_{2}  cos  α  –  F  –  T_{1}  cos  β = 0

T_{2} cos α  –  F = 0                    ①

Resolving vertically (\uparrow )

T_{1}  sin  β  +  T_{1}  sin  β  –  T_{2}  sin  α  –  W = 0

T_{1} sin β – T_{2}  sin  α  –  W = 0                ②

iii) When F = 50 and α = 45° equation ① gives

\begin{matrix}T_{2} cos 45° = 50    &    \longleftarrow       \boxed{ \text{ This tells you that } T_{2} \text{ is } \frac{50}{cos  45°} \text{ but you don’t need to work it out because }cos  45° = sin  45°.}\end{matrix}

⇒ T_{2} sin 45° = 50

Substituting in ② gives                      2T_{1} sin β  – 50 – W = 0

So when W = 100 and β = 75°                  2T_{1} sin 75° = 150

T_{1} = \frac{150}{2 sin  75°}

The tension in Brian’s rope is 77.65 N = 78 N (to the nearest N).

iv) When the wind blows harder, F increases. Given that all the angles remain unchanged, Eric will have to pull harder so the vertical component of T_{2} will increase. This means that T_{1} must increase and Brian must pull harder.