Statics of rigid bodies

In a nutshell

Sometimes it is necessary to consider the rotational forces acting on a body (e.g. a rod or a ladder). In these cases, you can consider the body to be a rigid body. If the rigid body is in static equilibrium, then:

• The body is stationary.
  
• The resultant force is zero.
• The resultant moment is zero. 

Equations

Variable definitions

Static rigid bodies

To solve problems involving static rigid bodies in equilibrium, consider that the resultant force on the body must be zero and the resultant moment must be zero.

PROCEDURE

Example 1

A uniform ladder has a length of $$l = 20\ m$$ and a mass of $$m= 3.5 \ kg$$. The ladder is resting against a smooth cylinder at $$P$$ as shown. The ladder is inclined $$\theta = 60 ^{\circ}$$. Find the value of the coefficient of friction $$\mu.$$​

As the diagram is already drawn, you have to resolve the forces into different components, parallel and perpendicular to the ground. You should add labels to your diagram as follows:

Resolve vertically to give:

$$R + N\cos(60) = 3.5 g \\R = 3.5g - N\cos(60)$$​​

Resolve horizontally to give:

$$F = N \sin (60)$$​​

Taking moments about $$A$$:

$$\begin{aligned}3.5g \times 10 \cos(60) &= N \times 15 \\\\N &= \dfrac{343}{30}\end{aligned}$$​​

Substitute into $$F=\mu R$$ to find $$\mu$$:

​$$\begin{aligned}F &= \mu R \\\\N \sin(60) &= \mu \times (3.5g - N \cos(60)) \\\\\dfrac{343}{30} \sin(60) &= \mu \times (3.5g - \dfrac{343}{30} \cos(60)) \\\\\mu &= \underline{0.346 \ (3 \ s.f.)}\end{aligned}$$​

Example 2

A uniform rod of mass $$m = 100 \ kg$$ rests with end $$A$$ on the floor. A string attached to the other end ($$B$$) keeps the rod in equilibrium. The string (under tension $$T$$) is perpendicular to the rod. Take into account that the rod is inclined at an angle of $$\theta = 45^{\circ}$$ and that it has a length of $$l = 10 \ m$$. Calculate the magnitudes of  $$T, \ R , \ \mu$$ and $$F$$.

First draw a diagram reflecting the information given in the question,

Next, resolve the forces into different components, parallel and perpendicular to the ground. You should add labels to your diagram as follows: 

Resolve vertically to give:

$$\begin{aligned}R + T \cos(45) &= 100g \\R&= 100g - T \cos(45)\end{aligned}$$​​

Resolve horizontally to give:

$$F = T \sin(45)$$​​

Taking moments about $$A$$:

$$\begin{aligned}100g \times 5 \cos(45) &= T \times 10 \\\\T&= \dfrac{100g \times 5 \cos(45)}{10} \\\\T&=346.48 \\\end{aligned}$$​​

Substitute in to the previous equations to find $$F$$ and $$R$$:

$$\begin{aligned}R&=100g - 346.48 \times \cos(45) \\R &=735 \\F &= 346.48 \times \sin(45) \\F&= 245 \end{aligned}$$​​

​​

Use $$F= \mu R$$ to find $$\mu$$:

​$$\begin{aligned}F &= \mu R \\245 &= \mu \times 735 \\\mu &= \frac 1 3\end{aligned}$$​​

​

$$\underline{T = 346 \ N, \ \ R = 735 \ N, \ \ F= 245 \ N, \ \ \mu = 0.333 \ (3 \ s.f.)}$$​​​