Static particles

​​In a nutshell

Static equilibrium is occurs when a body is stationary, and the resultant force acting on it is zero. It is very important to draw a diagram, showing all the forces involved, in order to calculate the missing forces. 

Equations

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Static particles

Solving problems with static particles

​​PROCEDURE

Example 1

The following diagram shows a particle in equilibrium under the action of three forces. Find the values of $$T$$ and $$\theta$$.

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As the diagram is already drawn, you can go directly to step 2, and resolve the forces into different components, parallel and perpendicular to the plane. You should add labels to your diagram as follows:

Set vertically upwards and horizontally to the right as positive, so that vertically downwards and horizontally to the left is negative.

Resolve parallel to the $$x$$ and $$y$$ axes:

$$\begin{cases}\text{x-axis:} \sum F_x = 0\implies T_x = 25 \ \ N\\\text{y-axis:} \sum F_y = 0\implies T_y = 42 \ \ N\end{cases}$$​​

Calculate the value of $$T$$ by using Pythagors' theorum:

$$T=\sqrt{25^2+42^2}=48.9 \ N$$​​

Now that you know the values of each component of $$T$$, you can find $$\theta$$​ by using some basic trigonometry: ​

​$$\begin{aligned}\tan \theta &= \dfrac{T_y}{T_x}\\\\\tan \theta &= \dfrac{42}{25}\\\\ \theta &= \tan^{-1}\Big(\dfrac{42}{25}\Big)\\\\\theta &= 59.2 \degree\end{aligned}$$​​

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Therefore $$\underline{T=48.9 \ N}$$ and $$\underline{\theta = 59.2\degree} \ (3 \ s.f.)$$

Example 2

Find the magnitudes of $$P$$ and $$Q$$ for the following system in equilibrium. 

As the force diagram is already drawn, you can go directly to step 2, and resolve the forces into different components: parallel and perpendicular to the plane. You should label your diagram as follows:

The system is in equilibrium, $$\sum F=0$$. Resolving parallel to the $$x$$-axis gives:

$$\begin{aligned}Q \cos (17) &= 10 \cos (47) \\\\Q &= \dfrac{10 \cos(47)}{\cos(17)} \\\\Q &= 7.1316...\end {aligned}$$​​

Resolve parallel to the $$y$$-axis to find $$P$$:

$$\begin{aligned}P + Q \sin(17) &= 10 \sin(47) \\P &= 10 \sin(47) - Q \sin(17) \\P &= 10 \sin(47) - 7.1316 \sin(17) \\P &= 5.2285\end{aligned}$$​​

Therefore, $$\underline{P=5.23 \ N}$$ and $$\underline{Q=7.13 \ N} \ (3 \ s.f.)$$​

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