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.

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

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 $$F_{A}=2.81 \mathrm{~N}$$.