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Application of forces

Static particles

Static particles

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Further kinematics

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Explainer Video

Tutor: Meera

Summary

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

DESCRIPTION

EQUATION

Definition of static equilibrium:

the sum of all the forces is zero.

​∑F=0\sum F = 0∑F=0​​


​

Static particles

Solving problems with static particles

​​PROCEDURE

1.

Draw a force diagram.

2.

Resolve the forces into different components.

3.

Form equations using ∑F=0\sum F = 0∑F=0 for each direction.

4.

Solve the equations. 


Example 1

The following diagram shows a particle in equilibrium under the action of three forces. Find the values of TTT and θ\thetaθ.

​

Maths; Application of forces; KS5 Year 13; Static particles


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:

Maths; Application of forces; KS5 Year 13; Static particles


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 xxx and yyy axes:

{x-axis:∑Fx=0 ⟹ Tx=25  Ny-axis:∑Fy=0 ⟹ Ty=42  N\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}{x-axis:∑Fx​=0⟹Tx​=25  Ny-axis:∑Fy​=0⟹Ty​=42  N​​​


Calculate the value of TTT by using Pythagors' theorum:

T=252+422=48.9 NT=\sqrt{25^2+42^2}=48.9 \ NT=252+422​=48.9 N​​


Now that you know the values of each component of TTT, you can find θ\thetaθ​ by using some basic trigonometry: ​

​tan⁡θ=TyTxtan⁡θ=4225θ=tan⁡−1(4225)θ=59.2°\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}tanθtanθθθ​=Tx​Ty​​=2542​=tan−1(2542​)=59.2°​​​

​​​

Therefore T=48.9 N‾\underline{T=48.9 \ N}T=48.9 N​ and θ=59.2°‾ (3 s.f.)\underline{\theta = 59.2\degree} \ (3 \ s.f.)θ=59.2°​ (3 s.f.)


Example 2

Find the magnitudes of PPP and QQQ for the following system in equilibrium. 

Maths; Application of forces; KS5 Year 13; Static particles


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:


Maths; Application of forces; KS5 Year 13; Static particles


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

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


Resolve parallel to the yyy-axis to find PPP:

P+Qsin⁡(17)=10sin⁡(47)P=10sin⁡(47)−Qsin⁡(17)P=10sin⁡(47)−7.1316sin⁡(17)P=5.2285\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}P+Qsin(17)PPP​=10sin(47)=10sin(47)−Qsin(17)=10sin(47)−7.1316sin(17)=5.2285​​​


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

​​


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FAQs - Frequently Asked Questions

What is the best way to start solving a problem of static particles?

The best way to start solving a problem of static particles is by drawing a force diagram.

What is the mathematical description of static equilibrium?

The sum of the forces in each direction must be zero.

When is static equilibrium achieved?

When the body is at rest and the resultant force acting on it is zero.

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