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 tension in each string.

Question

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 tension in each string.

Accepted Answer

The force diagram for this situation is given below.

Method 1: Resolving forces

Vertically ([latex]\uparrow[/latex]) : [latex]T_{1} sin 30° + T_{2} sin 45° - 10g = 0[/latex]

0.5 [latex]T_{1}[/latex] + 0.707 [latex]T_{2}[/latex] = 98 ①

Horizontally (→) : [latex]- T_{1} cos 30° + T_{2} cos 45° = 0[/latex]

-0.866 [latex]T_{1}[/latex] + 0.707 [latex]T_{2}[/latex] = 0 ②

Subtracting ② from ① 1.366 [latex]T_{1}[/latex] = 98

[latex]T_{1}[/latex] = 71.74

Back substitution gives [latex]T_{2}[/latex] = 87.87

The tensions are 71.7 N and 87.9 N (to 1 d.p.).

Method 2: Triangle of forces

Since the three forces are in equilibrium they can be represented by the sides of a triangle taken in order.

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