Dynamics
dynamics, branch of physical science and subdivision of
mechanics that is concerned with the motion of material
objects in relation to the physical factors that affect
them: force, mass, momentum, energy.
What are the 2 dynamics?
The study of dynamics falls under two
categories: linear and rotational.
https://www.youtube.com/watch?v=iAMxWvyykRU
Friction, force that resists the sliding or rolling of one
solid object over another. Frictional forces provide the
traction needed to walk without slipping
Example:
A boy pulls a sled mass of 20 kg horizontally over a surface with a coefficient of friction of
0.20. it takes him 10 minutes to pull the sled 100 m . What is the average power output
over these 10 minutes?
𝑊 = 20 𝑘𝑔
𝑁
𝐹
𝐹𝑅
𝐹𝑅
𝑊 = 10 𝑘𝑔
𝜇 = .20
𝑁
𝑑 = 10 𝑚
𝑇 = 10 𝑚𝑖𝑛.
𝑊𝑜𝑟𝑘 = 𝐹𝑑
𝑊𝑜𝑟𝑘 = 4 𝑘𝑔𝑥9.81 𝑚 𝑠 2
𝐹
100 𝑚 = 3924 𝐽
𝑤𝑜𝑟𝑘
3924 𝐽
𝑃=
=
6.54 𝑤𝑎𝑡𝑡𝑠
𝑡𝑖𝑚𝑒 (10 min 𝑥 60 𝑠𝑒𝑐 )
1 𝑚𝑖𝑛.
𝑣
𝑁 = 𝑊 = 20 𝑘𝑔
𝐹𝑅 = 𝜇𝑁 = 0.20 20 𝑘𝑔 = 4 𝑘𝑔
𝐻
𝐹 = 𝐹𝑅 = 4 𝑘𝑔
Example:
A 50 kg box is pushed across a horizontal floor by a horizontal force of 180 N. The
coefficient of sliding friction is 0.3. find the work done in pushing the box 20 m.
𝑊 = 50 𝑘𝑔
𝐻 = 𝐹 − 𝐹𝑅 = 𝐹𝑛𝑒𝑡
𝐹 = 180 𝑁
𝐹𝑅
𝐹 − 𝐹𝑅 = 𝐹𝑛𝑒𝑡
𝜇 = 0.30
𝑑 = 20 𝑚
𝑁
180 𝑁 − 147.15 𝑁 = 𝐹𝑛𝑒𝑡
𝐹𝑛𝑒𝑡 = 32.85 𝑁
𝑣=0
𝑁
𝐹𝑅
𝐹
𝑊 = 50 𝑘𝑔
𝑁 = 𝑊 = 50 𝑘𝑔𝑥9.81 𝑚 𝑠 2
𝑁 = 490.5 𝑁
𝐹𝑅 = 𝜇𝑁 = 0.30 490.5 𝑁
𝐹𝑅 = 147.15 𝑁
𝑤𝑜𝑟𝑘 = 𝐹𝑑
𝑤𝑜𝑟𝑘 = 32.85 𝑁 20 𝑚 = 657 𝐽
Example:
A 50 kg box is pushed across a horizontal floor by a force of 180 N @ 30𝑜 with the horizontal. The
coefficient of sliding friction is 0.3. find the work done in pushing the box 20 m.
𝑊 = 50 𝑘𝑔
𝐹 = 180 𝑁
30
𝐻 = 𝐹 − 𝐹𝑅 = 𝐹𝑛𝑒𝑡
𝑜
𝐹𝑅
𝐹𝑥 − 𝐹𝑅 = 𝐹𝑛𝑒𝑡
𝜇 = 0.30
𝑑 = 20 𝑚
𝑁
180 𝑁𝑐𝑜𝑠30 − 174.15 𝑁 = 𝐹𝑛𝑒𝑡
𝐹𝑛𝑒𝑡 = −18.26 𝑁
𝑣=0
𝑁
𝐹𝑅
𝑁 = 𝑊 + 𝐹𝑦
𝐹𝑥
30𝑜
𝑊 = 50 𝑘𝑔
𝐹𝑦
𝐹
= 50 𝑘𝑔𝑥9.81 𝑚 𝑠 2 + 180 𝑁𝑠𝑖𝑛30
𝑁 = 580.5 𝑁
𝐹𝑅 = 𝜇𝑁 = 0.30 580.5𝑁
𝐹𝑅 = 174.15𝑁
𝑤𝑜𝑟𝑘 = 𝑛𝑜 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒; 𝐹𝑟 > 𝐹𝑥
Example:
A 50 kg box is pulled across a horizontal floor by a force of 180 N @ 30𝑜 with the horizontal. The
coefficient of sliding friction is 0.3. find the work done in pushing the box 20 m.
𝑊 = 50 𝑘𝑔
𝐹 = 180 𝑁
𝐻 = 𝐹 − 𝐹𝑅 = 𝐹𝑛𝑒𝑡
30𝑜
𝐹𝑅
𝐹𝑥 − 𝐹𝑅 = 𝐹𝑛𝑒𝑡
𝜇 = 0.30
180 𝑁𝑐𝑜𝑠30 − 120.15𝑁 = 𝐹𝑛𝑒𝑡
𝑑 = 20 𝑚
𝑁
𝐹𝑛𝑒𝑡 = 35.73457268 𝑁
𝑁
𝐹𝑅
𝑣=0
𝐹
30
𝑜
𝐹𝑥
𝑊 = 50 𝑘𝑔
𝐹𝑦
𝑁 + 𝐹𝑦 = 𝑊
𝑤𝑜𝑟𝑘 = 𝐹𝑑
𝑤𝑜𝑟𝑘 = 35.73457268 𝑁 20 𝑚 = 714.6914 𝐽
𝑁 = 50 𝑘𝑔𝑥9.81 𝑚 𝑠 2 − 180 𝑁𝑠𝑖𝑛30
𝑁 = 400.5 𝑁
𝐹𝑅 = 𝜇𝑁 = 0.30 400.5𝑁
𝐹𝑅 = 120.15𝑁
A 500 lbs block is in contact with 45𝑜 incline. The coefficient of static friction is 0.5. find the horizontal force
necessary just to move the block up.
𝑁
𝑊
45𝑜
𝐹𝑅
𝐹
45𝑜
45𝑜
45𝑜
45𝑜
𝑁
𝑊 = 500 𝑙𝑏𝑠
𝐹
𝐻=0
𝐹𝑐𝑜𝑠45 − 𝑊𝑠𝑖𝑛45 − 𝐹𝑟 = 0
𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝐹𝑅
𝑣=0
𝑁 − 𝑊𝑐𝑜𝑠45 − 𝐹𝑠𝑖𝑛45 = 0
𝑁 − 500 𝑙𝑏𝑠 𝑐𝑜𝑠45 − 𝐹 𝑠𝑖𝑛45 = 0
𝑁 = 500 𝑙𝑏𝑠 𝑐𝑜𝑠45 + 𝐹 𝑠𝑖𝑛45
𝐹𝑐𝑜𝑠45 − 𝑊𝑠𝑖𝑛45 − 𝐹𝑟 = 0
𝐹𝑐𝑜𝑠45 − 500 𝑙𝑏𝑠 sin 45 − 𝜇𝑁 = 0
𝐹𝑐𝑜𝑠45 − 500 𝑙𝑏𝑠 sin 45 − 0.5(500 𝑙𝑏𝑠 𝑐𝑜𝑠45 + 𝐹 𝑠𝑖𝑛45 ) = 0
𝐹𝑐𝑜𝑠45 − 0.5(500 𝑙𝑏𝑠 𝑐𝑜𝑠45 + 𝐹 𝑠𝑖𝑛45 ) = 500 𝑙𝑏𝑠 sin 45
0.3535 𝐹 = 530.33 𝑙𝑏𝑠
𝐹 = 1500 𝑙𝑏𝑠
EXAMPLE:
A 100 lb. crate is initially stationary and is placed along an inclined
planed making an angle of 30𝑂 with the horizontal. The coefficient of
friction between the crate and the inclined surface are 𝜇𝑠 =
0.2 𝑎𝑛𝑑 𝜇𝑘 = 0.16.
A. If a horizontal force of 90 lbs is applied at the crate, determine how
far the crate moves from its position in 2 sec.
B. What is the acceleration of the block.
C. If the horizontal force applied to the create is only 30 lbs., how far
will the create moves from its initial position in 2 sec.