The picture shows three men involved in moving a packing case up to the top floor of a warehouse. Brian is pulling on a rope which passes round smooth pulleys at X and Y and is then secured to the point Z at the end of the loading beam. The wind is blowing directly towards the building. To

Question

The picture shows three men involved in moving a packing case up to the top floor of a warehouse. Brian is pulling on a rope which passes round smooth pulleys at X and Y and is then secured to the point Z at the end of the loading beam.

The wind is blowing directly towards the building. To counteract this, Eric is pulling on another rope, attached to the packing case at P, with just enough force and in the right direction to keep the packing case central between X and Z.

At the time of the picture the men are holding the packing case motionless.
i) Draw a diagram showing all the forces acting on the packing case using [latex]T_{1}[/latex] and [latex]T_{2}[/latex] for the tensions in Brian and Eric’s ropes, respectively.
ii) Write down equations for the horizontal and vertical equilibrium of the packing
case.

In one particular situation, W = 200, F = 50 , α = 45° and β = 75° .

iii) Find the tension [latex]T_{1}[/latex] .
iv) Explain why Brian has to pull harder if the wind blows stronger.

Accepted Answer

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

ii) Equilibrium equations

Resolving horizontally ( → )

[latex]T_{1} cos β + T_{2} cos α - F - T_{1} cos β = 0[/latex]

[latex]T_{2}[/latex] cos α - F = 0 ①

Resolving vertically ([latex]\uparrow [/latex] )

[latex]T_{1} sin β + T_{1} sin β - T_{2} sin α - W = 0[/latex]

[latex]T_{1}[/latex] sin β - [latex]T_{2}[/latex] sin α - W = 0 ②

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

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

⇒ [latex]T_{2}[/latex] sin 45° = 50

Substituting in ② gives 2[latex]T_{1}[/latex] sin β - 50 - W = 0

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

[latex]T_{1} = \frac{150}{2 sin 75°}[/latex]

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 [latex]T_{2}[/latex] will increase. This means that [latex]T_{1}[/latex] must increase and Brian must pull harder.

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

Related Answered Questions

Verified Answer:

Verified Answer:

Verified Answer:

This example illustrates two methods for solving problems involving forces in equilibrium. With experience, you will find it easier to judge which method is best for a particular problem. A sign of mass 10 kg is to be suspended by two strings arranged as shown in the diagram opposite. Find the ...

Verified Answer:

A brick of mass 3 kg is at rest on a rough plane inclined at an angle of 30° to the horizontal. Find the friction force F N, and the normal reaction R N of the plane on the brick. ...

Verified Answer: