# Rigid Bodies and Friction

## Rigid Bodies and Friction Revision

**Reactions Can Have Components**

- If a
**rod**is connected to a**plane**by a**hinge**or**pivot**, and there are**forces that are not parallel**to the**rod**acting on the**rod**, then the**reaction force**on the**rod**at the**wall**will**not be perpendicular**to the**wall**. We deal with this by**resolving it into parallel and perpendicular components**(i.e. components in the direction of the**rod**and in the direction of the**wall**).

- If the
**rod**is not attached to the**wall**, but is instead**held against the wall by friction**, the**friction replaces the vertical component of the reaction force**, and we only get the component**perpendicular to the****wall**, like usual.

- Sometimes, a
**rod**will be positioned against**two surfaces**(such as a**ladder**that is on both a**floor**and a**wall**– a**very common exam question**). There could be**friction on none, one, or two of the surfaces**. The question should indicate which surfaces are rough and which surfaces are smooth.

- Questions can also be asked about
**bodies**leaning against**surfaces**that are**neither horizontal nor vertical**. As ever, just place the reaction force**perpendicular to the surface**and the friction force**parallel to the surface**, and you are good to go.

- If a
**rod**has a support that it is resting on along its length, then the**reaction force is perpendicular**to the**rod**.

- Finally, some questions ask you to
**find a range for the coefficient of friction**, \mu. To do this,**treat the system as if it were in limiting equilibrium**, but when you come to work out the coefficient of friction, replace F=\mu R with F\leq\mu R.

**Example 1: Rod Held by a Pivot**

A **horizontal rod** is held against a **vertical wall** by a **pivot**. The weight of the rod is 22\text{ N}, and it is held by a bar at 60\degree, providing a tension of 50\text{ N}. Find the **horizontal and vertical components** of the** reaction force** of the **rod** against the **wall**.

**[4 marks]**

First, draw a diagram.

Resolve vertically:

50\cos(60)=R_{v}+22

25=R_{v}+22

R_{v}=3 \text{ N}

Resolve horizontally:

50\sin(60)=R_{h}

R_{h}=25\sqrt{3}

R_{h}=43.3 \text{ N}

**Example 2: Ladder**

A **ladder** is stood upon a **rough horizontal floor** and leaning against a **smooth vertical wall**. Given that it has a **mass** of 12\text{ kg}, and the **reaction force** from the **vertical wall** is 29.4\text{ N}, find the **coefficient of friction** of the **floor**.

**[3 marks]**

First draw a diagram.

Resolve vertically:

R=12gResolve horizontally:

F=29.4\text{ N}

F \leq \mu R

29.4 \leq 12g\mu

29.4 \leq 117.6\mu

\mu\geq \dfrac{29.4}{117.6}

\mu \geq 0.25

## Rigid Bodies and Friction Example Questions

**Question 1:** A 30\text{ cm} shelf is propped up by a support beam, as below. Given that the support beam can supply a maximum of 300\text{ N}, could the shelf support two \text{7.5 kg} trophies at distances 15\text{ cm} and 25\text{ cm}?

Also, calculate the horizontal and vertical reaction forces at the point where the shelf meets the wall.

Assume that the shelf is weightless.

**[6 marks]**

Take the moment about the point where the shelf meets the wall.

Clockwise: (0.15 \times 7.5 \times 9.8) + (0.25 \times 7.5 \times 9.8) = 0.4 \times 7.5 \times 9.8 = 29.4\text{ Nm}

Anticlockwise: The maximum perpendicular force is 0.15 \times 300\cos 30° = 38.97\text{ Nm (to } 2\text{ dp)}

The force from the support beam exceeds the force from the trophies, so the shelf will stay up.

For the entire system, vertically:

R_{Vwall} + T_{1}\cos 30° = 2 \times 7.5 \times 9.8 = 147

Since we know that

T_{1} = \dfrac{29.4}{0.15\cos 30°} = 226.3, we have R_{Vwall} = -49\text{ N}

so this reactionary force acts downwards at the wall.

For the entire system, horizontally:

R_{Hwall} = T_1 \sin 30° = 113.16\text{ N}

**Question 2:** An art piece can be modelled as a uniform rod weighing 5\text{ kg}, held up by two large, weightless pipes.

Given that the piece is in equilibrium, find T_1 and T_2, using the moments technique.

**[4 marks]**

Resolving horizontally, we have

T_1\cos 26.6 = T_2\cos 58

Rearrange this to get

T_1 = T_2\dfrac{\cos 58}{\cos 26.6}

Using moments about the left end of the rod:

Clockwise: 2.5 \times 5 \times 9.8 = 122.5\text{ Nm}

Anticlockwise: T_1\sin 26.6 + 4.5T_2\sin 58

122.5 = T_1\sin 26.6 + 4.5T_2\sin 58\\

122.5 = (T_2 \times \tan 26.6 \times \cos 58) + (4.5T_2\sin 58)\\

T_2 = \dfrac{122.5}{(\tan 26.6 \times \cos 58) + 4.5\sin 58} = 30.01\text{ N}\\

T_1 = 17.79\text{ N}

**Question 3: **Consider a ladder placed on a rough floor, at an angle of 60\degree to the horizontal. The ladder rests on a vertical wall, and the coefficient of friction of both the floor and the wall is 0.5. The ladder has a mass of 20\text{ kg}, which is evenly distributed across its 1\text{ m} length. By taking moments about the base, or otherwise, find the reaction forces from both the floor and the wall.

**[5 marks]**

Draw a diagram.

Evenly distributed so the weight acts in the middle, 0.5\text{ m} from the base.

Take moments about the base:

R_{2}\sin(60)=0.5R_{2}\cos(60)+0.5\times20g\cos(60)

\dfrac{\sqrt{3}}{2}R_{2}=\dfrac{1}{4}R_{2}+5g

\left(\dfrac{\sqrt{3}}{2}-\dfrac{1}{4}\right)R_{2}=5g

R_{2}=\dfrac{5g}{\dfrac{\sqrt{3}}{2}-\dfrac{1}{4}}

R_{2}=79.5\text{ N}

Resolve forces horizontally.

R_{2}=0.5R_{1}

0.5R_{1}=79.5\text{ N}

R_{1}=159.1\text{ N}

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