Rigid Bodies and Friction

Rigid Bodies and Friction

A LevelAQAEdexcelOCREdexcel 2022OCR 2022

Rigid Bodies

Rigid bodies such as rods interact with walls and surfaces in ways that you might not expect. There are several key facts you need to know for these questions.

Make sure you are happy with the following topics before continuing.

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.
A LevelAQAEdexcelOCR

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}

A LevelAQAEdexcelOCR

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=12g

Resolve horizontally:

F=29.4\text{ N}

F=\mu R

29.4=12g\mu

29.4=117.6\mu

\mu=\dfrac{29.4}{117.6}

\mu=0.25

A LevelAQAEdexcelOCR

Example Questions

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}

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}

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}

Related Topics

MME

Moments

A Level
MME

Friction

A Level

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

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