Modelling with statics

In a nutshell

When modelling problems that involve forces such as tension and weight, it is very useful to draw a force diagram. It is possible to find the values of the unknown forces by using the fact that $$\sum F=0$$ and resolving forces into perpendicular components. You can resolve horizontally and vertically as you have already learnt or it can be easier to resolve parallel or perpendicular to a particular plane or surface.​

Modelling with statics

Procedure

Note: Remember that when solving problems that involve tension, the tension in the string will be the same on both sides if it is one string. 

Example 1

A particle of mass $$m= 2 \ kg$$ rests on a smooth inclined plane, which makes an angle of $$\theta = 20^{\circ}$$ with the horizontal. The particle is maintained in equilibrium by a force $$P$$, as shown in the diagram. Find the value of $$P$$.

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:

Note which directions to make positive. In this case, parallel up the plane and perpendicular upwards are considered positive.

Resolve forces and use $$\sum F= 0$$ to obtain:

$$\begin{cases}\begin{aligned}\text{parallel to the plane: }& P - W_x = 0 \\\text{perpendicular to the plane: } & R - W_y = 0\end{aligned}\end{cases}$$​​

So:

$$P = W_x = mg\sin\theta = 10 g \sin 20^{\circ}$$​​

Therefore, $$\underline{P = 33.5 \ N \ (3 \ s.f.)}$$.

Example 2

A particle of mass $$m=10\ kg$$ hangs in equilibrium, suspended by two light inextensible strings, as shown below. Find the tension in each of the strings. 

As the diagram is already drawn, you can go directly to step 2, and resolve the forces into horizontal and vertical components. You should add labels to your diagram as follows. 

Note vertically downwards and horizontally to the right as positive. 

Resolve the forces into components and use $$\sum F = 0$$ to obtain:

​​$$\begin{aligned}T_1 \cos(60) &= T_2 \cos(30) \qquad &\textcircled{1}\\T_1 \sin(6)+T_2\sin(30)&=98 \qquad &\textcircled{2}\end{aligned}$$​

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Solve simultaneously to give:

​$$\begin{aligned}T_1=T_2 \dfrac{\cos(30)}{\cos(60)} \qquad \textcircled{1}\\\\T_2 \dfrac{\cos(30)}{\cos(60)} \times \sin(60) + T_2 \sin(30) &=98 \qquad \textcircled{2}\\\\T_2 \Big(\dfrac{\cos(30)}{\cos(60)} \times \sin(60) + \sin(30)\Big) &=98 \\\\T_2&= \underline{49 \ N} \\\\T_1=49 \times \dfrac{\cos(30)}{\cos(60)} \\\\T_1 = 49 \sqrt{3} \\T_1 = \underline{84.9 \ N \ (3 \ s.f.)}\end{aligned}$$​​

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