1. Torque Rotational analogue of Force Must contain Use door example
Force - larger force, more twisting
Distance – larger lever arm, more twisting
“Perpendicularness” – more right angles, more twisting
Direction – can twist 2 directions
Use door example

2. Torque Definition 𝜏 =𝑟 𝐹 𝑠𝑖𝑛𝜃 Force times distance to rotation point
Sin(θ) gives “perpendicularness”
Can be F┴ times d
Can be d┴ times F
Figure 8-13
Causes rotation convention CCW +, CW –
Units meter-newtons, NOT joules
Usually consider vector, with direction
“cross” product
“right hand rule”
absolute value sign

3. 2 ways to multiply vectors
Dot Product
𝑊=𝐹∙𝑑=𝐹𝑑 𝑐𝑜𝑠𝜃
Example - work and energy
Extracts amount 2 vectors are inline
Result scalar
Cross product
𝜏=𝑟×𝐹 𝜏 =𝑟 𝐹 𝑠𝑖𝑛𝜃
Example – torque
Extracts amount 2 vectors are perpendicular
Result vector (along axis of rotation)*
*we’re not going to use much

4. Example 8-8 Arm at right angles Arm at 60° Axis at elbow Bicep at 5 cm
𝜏 =.05 𝑚 700 𝑁 𝑠𝑖𝑛90
𝜏 =35 𝑚 𝑁
Arm at 60°
𝜏 =.05 𝑚 700 𝑁 𝑠𝑖𝑛60
Use perpendicular force
𝑟 𝐹 ┴ =(.05 𝑚)(700 𝑁 𝑠𝑖𝑛60)
𝑟𝐹 ┴ =(.05 𝑚)(700 𝑁 𝑐𝑜𝑠30)
Use perpendicular moment arm
𝑟 ┴ 𝐹=(.05 𝑚 𝑠𝑖𝑛60)(700 𝑁)
Either way 𝜏 =30 𝑚 𝑁

5. Example 8-9 Torque on inner wheel (r =0.3 m)
𝜏 =.3 𝑚 50 𝑁 𝑠𝑖𝑛90
𝜏 =15 𝑚 𝑁
CounterClockwise (show direction)
Positive
Torque on outer wheel (r =0.5 m)
𝜏 =.5 𝑚 50 𝑁 𝑠𝑖𝑛60 or
𝜏 =.5 𝑚 50 𝑁 𝑐𝑜𝑠30
𝜏 =−21.7 𝑚 𝑁
Clockwise (show direction)
Negative
Total 𝜏 =15−21.7=−6.7 𝑚 𝑁

6. Problem 27 Calculate Net Torque Left-hand torque negative
Right-hand torque positive
ϴ = 90 for both
Net Torque

7. Problem 28 a) Calculate force 28 cm away
b) Calculate force on 6 sided nut

8. Problem 25