Problem

The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and hel...

The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in place by the spring which has an unstretched length of 200 mm. Determine the magnitude of the resultant force at A and the normal force on the peg at B when the lever is in the position shown.

Step-by-Step Solution

Solution 1

Draw the free body diagram of the lever.

417-5-51P_fig1

Consider triangle PQR and calculate the length of the spring and the angle .

417-5-51P_fig2

Calculate the length of the spring by applying cosine rule to the triangle \(P Q R\).

\(l^{2}=300^{2}+400^{2}-2 \times 300 \times 400 \cos 150^{\circ}\)

\(l=\sqrt{300^{2}+400^{2}-2 \times 300 \times 400 \cos 150^{\circ}}\)

\(=676.64 \mathrm{~mm}\)

Calculate the angle \(\phi\) by applying sine rule to the triangle \(\mathrm{PQR}\).

\(\frac{\sin 150}{l}=\frac{\sin \phi}{300}\)

Substitute \(676.64 \mathrm{~mm}\) for \(l\).

\(\frac{\sin 150^{\circ}}{676.64}=\frac{\sin \phi}{300}\)

\(\sin \phi=\frac{300}{676.64} \sin 150^{\circ}\)

\(\phi=12.8^{\circ}\)

Calculate the spring force by using the equation,

$$ F_{S}=K(l-\delta) \ldots \ldots(1) $$

Here, \(F_{S}\) is the spring force, \(K\) is the stiffness of the spring, \(l\) is the length of the spring and \(\delta\) is the unstretched length of the spring.

Substitute \(5 \mathrm{~N} / \mathrm{m}\) for \(K, 0.67664 \mathrm{~m}\) for \(l\) and \(0.2 \mathrm{~m}\) for \(\delta\).

$$ \begin{aligned} F_{S} &=5 \times(0.67664-0.2) \\ &=2.38 \mathrm{~N} \end{aligned} $$

Calculate the algebraic sum of the moments of all the forces about point \(A .\)

\(\left(N_{B} \times 100\right)-\left(F_{s} \sin \phi \times 400\right)=0\)

Substitute \(2.38 \mathrm{~N}\) for \(F_{s}\) and \(12.8^{\circ}\) for \(\phi\).

\(\left(N_{B} \times 100\right)-\left(2.38 \sin 12.8^{\circ} \times 400\right)=0\)

\(N_{B}=\frac{\left(2.38 \sin 12.8^{\circ} \times 400\right)}{100}\)

\(=2.11 \mathrm{~N}\)

Therefore, the normal force on the peg at \(B\) is \(N_{B}=2.11 \mathrm{~N}\).

Apply equilibrium equation to the lever along the horizontal direction.

\(\sum F_{x}=0\)

\(A_{x}-F_{s} \cos \phi=0\)

Substitute \(2.38 \mathrm{~N}\) for \(F_{s}\) and \(12.8^{\circ}\) for \(\phi\).

\(A_{x}-F_{s} \cos \phi=0\)

\(\begin{aligned} A_{x} &=F_{S} \cos \phi \\ &=2.38 \cos 12.8^{\circ} \\ &=2.32 \mathrm{~N} \end{aligned}\)

Apply equilibrium equation to the lever along the vertical direction.

\(\sum F_{y}=0\)

\(A_{y}+N_{B}-F_{S} \sin \phi=0\)

Substitute \(2.38 \mathrm{~N}\) for \(F_{s}, 2.11 \mathrm{~N}\) for \(N_{B}\) and \(12.8^{\circ}\) for \(\phi\).

\(A_{y}+N_{B}-F_{S} \sin \phi=0\)

\(\begin{aligned} A_{y} &=F_{s} \sin \phi-N_{B} \\ &=2.38 \sin 12.8^{\circ}-2.11 \\ &=-1.58 \mathrm{~N} \end{aligned}\)

Calculate the resultant force at \(\mathrm{A}\) by using the following equation:

$$ F_{A}=\sqrt{A_{x}^{2}+A_{y}^{2}} $$

Here, \(F_{A}\) is the resultant force at \(\mathrm{A}\).

Substitute \(2.32 \mathrm{~N}\) for \(A_{x}\) and \(-1.58 \mathrm{~N}\) for \(A_{y}\).

$$ \begin{aligned} F_{A} &=\sqrt{2.32^{2}+(-1.58)^{2}} \\ &=2.81 \mathrm{~N} \end{aligned} $$

Therefore, the resultant force at \(\mathrm{A}\) is \(F_{A}=2.81 \mathrm{~N}\).

Add your Solution
Textbook Solutions and Answers Search