Figure 7.32 shows a model for a step-ladder standing on a smooth horizontal floor.

i) Draw a diagram to show the forces acting on both rods AB and BC.
ii) Explain why the internal forces in the hinge at B are horizontal.
iii) Calculate the tension in the string.

A woman of mass 56 kg stands on a step 0.5 m from A. Calculate
iv) the new reaction at C
v) the tension in the string and the magnitude of the reaction at B.

Question

Figure 7.32 shows a model for a step-ladder standing on a smooth horizontal floor.

i) Draw a diagram to show the forces acting on both rods AB and BC.
ii) Explain why the internal forces in the hinge at B are horizontal.
iii) Calculate the tension in the string.

A woman of mass 56 kg stands on a step 0.5 m from A. Calculate
iv) the new reaction at C
v) the tension in the string and the magnitude of the reaction at B.

Accepted Answer

i) Figure 7.33 shows the forces acting when the ground is smooth.

ii) By Newton’s third law, the reactions at B are equal and opposite. Also by the symmetry of the step-ladder and the forces, the vertical components are equal in magnitude and direction. Both conditions can be satisfied only if Y = 0 and [latex]R_{1} = R_{2}[/latex].
iii) The two parts of the ladder can be treated separately.
Resolving horizontally for AB or BC ⇒ X = T
Taking moments about C for BC
⇒ 10g × 1 cos 70° + T × 1 sin 70° = X × 2 sin 70°
Substituting for X 10g cos 70° = 2T sin 70° - T sin 70°

⇒ T = [latex]\frac{10g cos 70°}{sin 70°}[/latex]

The tension in the string is 35.7 N.
iv) When a woman stands on a step, the forces are no longer symmetrical so

Y ≠ 0 and [latex]R_{1} ≠ R_{2}[/latex]

There are now several possible ways forward. You can treat the ladder as a whole or each part separately but it is best to try to avoid writing down too many equations involving a lot of unknowns.
Take moments about A for the whole system:

[latex]R_{2}[/latex] × 4 cos 70° = 10g × 3 cos 70° + 10g × 1 cos 70° + 56g × 0.5 cos 70°

[latex]\begin{matrix}R_{2} = 30g + 10g + 28g &\longleftarrow \ \end{matrix} \begin{matrix} \boxed{ ext{Divide by cos 70°}} \end{matrix}[/latex]

[latex]R_{2} = [/latex]17g

The new reaction at C is 167 N.

v) Take moments about B for BC:

[latex]R_{2} [/latex] × 2 cos 70° = 10g × 1 cos 70° + T × 1 sin 70°

Substituting from ① ⇒ (34g - 10g) cos 70° = T sin 70°

T = [latex]\frac{24g cos 70°}{sin 70°}[/latex]

The tension in the string is 85.6 N.

Resolve vertically for BC: [latex]R_{2} [/latex] = 10g + Y

⇒ Y = 7g = 68.6

horizontally: X = T = 85.6

The reaction at B has magnitude [latex]\sqrt{(85.6² + 68.6²)} = 110 N[/latex]
___________________________________________
Notice that the extra vertical force due to the woman’s weight has an effect on all the forces including those that are horizontal. You cannot assume that any of the reactions remain as they were.

Question 7.6: Figure 7.32 shows a model for a step-ladder standing on a sm...

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