using newton’s laws 298 summer 19/lectures/… · summer 2018 prof. sergio b. mendes 1 chapter 5...

1Prof. Sergio B. MendesSummer 2018

Chapter 5 of Essential University Physics, Richard Wolfson, 3rd Edition

Using Newton’s Laws


Example 5.1

𝑎𝑎 = ? ?

𝑚𝑚 = 65 𝑘𝑘𝑘𝑘

𝜃𝜃 = 32°

𝑛𝑛 = ? ?


𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚 𝒂𝒂

𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝒏𝒏 + 𝑭𝑭𝒈𝒈

𝒏𝒏 + 𝑭𝑭𝒈𝒈 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 0 + 𝑚𝑚 𝑘𝑘 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 = 𝑚𝑚 𝑎𝑎𝑥𝑥𝑦𝑦: 𝑛𝑛 −𝑚𝑚 𝑘𝑘 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑚𝑚 𝑎𝑎𝑦𝑦 = 0

𝑎𝑎𝑥𝑥 = 𝑘𝑘 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃

𝑛𝑛 = 𝑚𝑚 𝑘𝑘 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃


Example 5.2


𝜃𝜃 = 22°

𝑡𝑡𝑡𝑡𝑛𝑛𝑠𝑠𝑠𝑠𝑐𝑐𝑛𝑛 𝑐𝑐𝑛𝑛 𝑡𝑡𝑎𝑎𝑐𝑐𝑒 𝑟𝑟𝑐𝑐𝑟𝑟𝑡𝑡 = ? ?

𝑎𝑎 = 0 𝑚𝑚/𝑠𝑠2



𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑻𝑻𝟏𝟏 + 𝑻𝑻𝟐𝟐 + 𝑭𝑭𝒈𝒈

𝑻𝑻𝟏𝟏 + 𝑻𝑻𝟐𝟐 + 𝑭𝑭𝒈𝒈 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 𝑇𝑇1 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 − 𝑇𝑇2 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑚𝑚 𝑎𝑎𝑥𝑥 = 0

𝑦𝑦: 𝑇𝑇1 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 + 𝑇𝑇2 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 −𝑚𝑚𝑘𝑘 = 𝑚𝑚 𝑎𝑎𝑦𝑦 = 0

𝑇𝑇1 = 𝑇𝑇𝟐𝟐 =𝑚𝑚 𝑘𝑘

2 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃


Example 5.3


𝜃𝜃 = 30°

𝑎𝑎 = 0 𝑚𝑚/𝑠𝑠2

𝐹𝐹ℎ = ? ?



𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝒏𝒏 + 𝑭𝑭𝒉𝒉 + 𝑭𝑭𝒈𝒈

𝒏𝒏 + 𝑭𝑭𝒉𝒉 + 𝑭𝑭𝒈𝒈 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 𝑛𝑛 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 − 𝐹𝐹ℎ = 𝑚𝑚 𝑎𝑎𝑥𝑥 = 0

𝑦𝑦: 𝑛𝑛 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 − 𝑚𝑚𝑘𝑘 = 𝑚𝑚 𝑎𝑎𝑦𝑦 = 0

𝐹𝐹ℎ = 𝑚𝑚𝑘𝑘 𝑡𝑡𝑎𝑎𝑛𝑛 𝜃𝜃


5.1Got It ?

A roofer's toolbox rests on an essentially frictionless metal roof with a 45 slope, secured by a horizontal rope as shown. Is the rope tension

a) greater thanb) less thanc) equal to the box's

weight

© 2016 Pearson Education, Inc.

A roofer's toolbox rests on an essentially frictionless metal roof with a 45 slope, secured by a horizontal rope as shown. Is the rope tension

a) greater thanb) less thanc) equal to the box's

weight

© 2016 Pearson Education, Inc.

Equal—but only because of the 45 slope. At larger angles, the tension would be greater than the weight; at smaller angles, less.


𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑟𝑟 𝒂𝒂𝒓𝒓

𝑻𝑻𝑳𝑳 + 𝑻𝑻𝑹𝑹 = 𝑚𝑚𝑟𝑟 𝒂𝒂𝒓𝒓

𝑎𝑎𝑟𝑟 ≅ 0

𝑇𝑇𝐿𝐿 ≅ 𝑇𝑇𝑟𝑟

𝑚𝑚𝑟𝑟 ≅ 0or


Example 5.4

𝑚𝑚𝑐𝑐 = 73 𝑘𝑘𝑘𝑘

𝑎𝑎𝑐𝑐 = ? ?

𝑚𝑚𝑟𝑟 = 940 𝑘𝑘𝑘𝑘

𝑡𝑡 = ? ?

∆𝑥𝑥 = 51 𝑚𝑚


𝒏𝒏 + 𝑻𝑻𝒓𝒓 + 𝑭𝑭𝒈𝒈,𝒓𝒓 = 𝑚𝑚𝑟𝑟 𝒂𝒂𝒓𝒓

𝑻𝑻𝒄𝒄 + 𝑭𝑭𝒈𝒈,𝒄𝒄 = 𝑚𝑚𝑐𝑐 𝒂𝒂𝒄𝒄

𝑎𝑎𝑟𝑟 = 𝑎𝑎𝑐𝑐 = 𝑎𝑎

𝑇𝑇𝑟𝑟 = 𝑚𝑚𝑟𝑟 𝑎𝑎𝑟𝑟𝑛𝑛 − 𝑚𝑚𝑟𝑟 𝑘𝑘 = 0

−𝑇𝑇𝑐𝑐 + 𝑚𝑚𝑐𝑐 𝑘𝑘 = 𝑚𝑚𝑐𝑐 𝑎𝑎𝑐𝑐

𝑇𝑇𝑟𝑟 = 𝑇𝑇𝑐𝑐 = 𝑇𝑇

𝑎𝑎 =𝑚𝑚𝑐𝑐 𝑘𝑘

𝑚𝑚𝑐𝑐 + 𝑚𝑚𝑟𝑟

𝑥𝑥:

𝑦𝑦:

𝑥𝑥:

𝑡𝑡 =2 ∆𝑥𝑥𝑎𝑎

rock

climber


Example 5.5

𝑣𝑣 = ? ?

𝑇𝑇 = ? ?

𝑚𝑚

𝜃𝜃

𝐿𝐿


𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑭𝑭𝒈𝒈 + 𝑻𝑻 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 𝑇𝑇 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑚𝑚 𝑎𝑎𝑥𝑥

𝑦𝑦: −𝑚𝑚𝑘𝑘 + 𝑇𝑇 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 = 𝑚𝑚 𝑎𝑎𝑦𝑦 = 0

𝑎𝑎𝑥𝑥 =𝑣𝑣2

𝑟𝑟=

𝑣𝑣2

𝐿𝐿 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃

𝑣𝑣 =𝑘𝑘 𝐿𝐿 𝑐𝑐𝑐𝑐𝑠𝑠2 𝜃𝜃𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃

𝑇𝑇 =𝑚𝑚 𝑘𝑘𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃

𝑇𝑇 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃𝑚𝑚

=


Example 5.6

𝑣𝑣 = 90𝑘𝑘𝑚𝑚𝑒

= 25𝑚𝑚𝑠𝑠

𝜃𝜃 = ? ?

𝑟𝑟 = 350 𝑚𝑚


𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑭𝑭𝒈𝒈 + 𝒏𝒏 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 𝑛𝑛 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 = 𝑚𝑚 𝑎𝑎𝑥𝑥

𝑦𝑦: −𝑚𝑚 𝑘𝑘 + 𝑛𝑛 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑚𝑚 𝑎𝑎𝑦𝑦 = 0

𝜃𝜃 = 𝑡𝑡𝑎𝑎𝑛𝑛−1𝑣𝑣2

𝑘𝑘 𝑟𝑟

𝑎𝑎𝑥𝑥 =𝑣𝑣2

𝑟𝑟

𝜃𝜃 = 10°


𝑣𝑣 = ? ?

𝑟𝑟 = 6.3 𝑚𝑚

Example 5.7


𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑭𝑭𝒈𝒈 + 𝒏𝒏 = 𝑚𝑚 𝒂𝒂

𝑦𝑦: 𝑚𝑚𝑘𝑘 + 𝑛𝑛 = 𝑚𝑚 𝑎𝑎𝑦𝑦

𝑎𝑎𝑦𝑦 =𝑣𝑣2

𝑟𝑟

𝑣𝑣𝑚𝑚𝑚𝑚𝑛𝑛 = 𝑟𝑟 𝑘𝑘𝑣𝑣 = 𝑟𝑟 𝑘𝑘 +𝑛𝑛𝑚𝑚

𝑣𝑣𝑚𝑚𝑚𝑚𝑛𝑛 = 7.9𝑚𝑚𝑠𝑠


Friction Force


Walking


𝑓𝑓𝑠𝑠 ≤ 𝜇𝜇𝑠𝑠 𝑛𝑛

𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑛𝑛

𝜇𝜇𝑠𝑠 > 𝜇𝜇𝑘𝑘

Static and Kinetic Friction

Static

Kinetic

𝑓𝑓𝑠𝑠, 𝑚𝑚𝑚𝑚𝑥𝑥 = 𝜇𝜇𝑠𝑠 𝑛𝑛


Example 5.10

𝜇𝜇𝑠𝑠 = 0.46 𝜃𝜃 = ? ?


𝑭𝑭𝒈𝒈 + 𝒏𝒏 + 𝒇𝒇𝒔𝒔 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 𝑚𝑚 𝑘𝑘 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 − 𝑓𝑓𝒔𝒔 = 0

𝑦𝑦: −𝑚𝑚 𝑘𝑘 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 + 𝑛𝑛 = 0

= 0

𝑓𝑓𝑠𝑠 = 𝑛𝑛 𝑡𝑡𝑎𝑎𝑛𝑛 𝜃𝜃

𝑓𝑓𝑠𝑠 ≤ 𝜇𝜇𝑠𝑠 𝑛𝑛𝑡𝑡𝑎𝑎𝑛𝑛 𝜃𝜃 ≤ 𝜇𝜇𝑠𝑠 𝜃𝜃 ≤ 𝑡𝑡𝑎𝑎𝑛𝑛−1 𝜇𝜇𝑠𝑠

𝜃𝜃 ≤ 𝑡𝑡𝑎𝑎𝑛𝑛−1 0.46 = 25°


Example 5.11

𝜇𝜇𝑘𝑘

𝑇𝑇 = ? ?

𝑚𝑚

𝜃𝜃

𝑣𝑣 = 𝑐𝑐𝑐𝑐𝑛𝑛𝑠𝑠𝑡𝑡𝑎𝑎𝑛𝑛𝑡𝑡


𝑭𝑭𝒈𝒈 + 𝒏𝒏 + 𝑻𝑻 + 𝒇𝒇𝒌𝒌 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 𝑇𝑇 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 − 𝑓𝑓𝑘𝑘 = 0

𝑦𝑦: −𝑚𝑚 𝑘𝑘 + 𝑛𝑛 + 𝑇𝑇 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃 = 0

= 0

𝑛𝑛 = 𝑚𝑚 𝑘𝑘 − 𝑇𝑇 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃

𝑇𝑇 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝑓𝑓𝑘𝑘

𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑛𝑛

𝑇𝑇 𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 = 𝜇𝜇𝑘𝑘 𝑚𝑚 𝑘𝑘 − 𝑇𝑇 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃

𝑇𝑇 =𝜇𝜇𝑘𝑘 𝑚𝑚 𝑘𝑘

𝑐𝑐𝑐𝑐𝑠𝑠 𝜃𝜃 + 𝜇𝜇𝑘𝑘 𝑠𝑠𝑠𝑠𝑛𝑛 𝜃𝜃


Example 5.9

𝑟𝑟 = 73 𝑚𝑚 𝑣𝑣𝑚𝑚𝑚𝑚𝑥𝑥 = ? ?𝜇𝜇𝑠𝑠 = 0.21

𝜇𝜇𝑠𝑠 = 0.88

𝑭𝑭𝒈𝒈 + 𝒏𝒏 + 𝒇𝒇𝒔𝒔 = 𝑚𝑚 𝒂𝒂

𝑥𝑥: 𝑓𝑓𝑠𝑠 = 𝑚𝑚 𝑎𝑎

𝑦𝑦: −𝑚𝑚 𝑘𝑘 + 𝑛𝑛 = 0

𝑓𝑓𝑠𝑠,𝑚𝑚𝑚𝑚𝑥𝑥 = 𝜇𝜇𝑠𝑠 𝑛𝑛

𝑣𝑣𝑚𝑚𝑚𝑚𝑥𝑥 = 𝑟𝑟 𝜇𝜇𝑠𝑠 𝑘𝑘

𝑎𝑎 =𝑣𝑣2

𝑟𝑟

𝜇𝜇𝑠𝑠 = 0.88𝑣𝑣𝑚𝑚𝑚𝑚𝑥𝑥 = 25 𝑚𝑚/𝑠𝑠

𝜇𝜇𝑠𝑠 = 0.21𝑣𝑣𝑚𝑚𝑚𝑚𝑥𝑥 = 12 𝑚𝑚/𝑠𝑠


Drag Forces

Terminal Velocity

Depends on:

• Instantaneous velocity

• Area opposing the motion

Air Drag

using newton’s laws 298 summer 19/lectures/… · summer 2018 prof. sergio b. mendes 1 chapter 5...

Documents