Newton’s Laws The Study of Dynamics. Introduction to Newton’s Laws Newton’s First Law.
Using Newton’s Laws 298 summer 19/lectures/… · Summer 2018 Prof. Sergio B. Mendes 1 Chapter 5...
Transcript of Using Newton’s Laws 298 summer 19/lectures/… · Summer 2018 Prof. Sergio B. Mendes 1 Chapter 5...
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1Prof. Sergio B. MendesSummer 2018
Chapter 5 of Essential University Physics, Richard Wolfson, 3rd Edition
Using Newton’s Laws
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2Prof. Sergio B. MendesSummer 2018
Example 5.1
𝑎𝑎 = ? ?
𝑚𝑚 = 65 𝑘𝑘𝑘𝑘
𝜃𝜃 = 32°
𝑛𝑛 = ? ?
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3Prof. Sergio B. MendesSummer 2018
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
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4Prof. Sergio B. MendesSummer 2018
Example 5.2
𝑚𝑚 = 17 𝑘𝑘𝑘𝑘
𝜃𝜃 = 22°
𝑡𝑡𝑡𝑡𝑛𝑛𝑠𝑠𝑠𝑠𝑐𝑐𝑛𝑛 𝑐𝑐𝑛𝑛 𝑡𝑡𝑎𝑎𝑐𝑐𝑒 𝑟𝑟𝑐𝑐𝑟𝑟𝑡𝑡 = ? ?
𝑎𝑎 = 0 𝑚𝑚/𝑠𝑠2
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5Prof. Sergio B. MendesSummer 2018
𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚 𝒂𝒂
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
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6Prof. Sergio B. MendesSummer 2018
Example 5.3
𝑚𝑚 = 62 𝑘𝑘𝑘𝑘
𝜃𝜃 = 30°
𝑎𝑎 = 0 𝑚𝑚/𝑠𝑠2
𝐹𝐹ℎ = ? ?
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7Prof. Sergio B. MendesSummer 2018
𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚 𝒂𝒂
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
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8Prof. Sergio B. MendesSummer 2018
5.1Got It ?
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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.
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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.
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11Prof. Sergio B. MendesSummer 2018
𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑟𝑟 𝒂𝒂𝒓𝒓
𝑻𝑻𝑳𝑳 + 𝑻𝑻𝑹𝑹 = 𝑚𝑚𝑟𝑟 𝒂𝒂𝒓𝒓
𝑎𝑎𝑟𝑟 ≅ 0
𝑇𝑇𝐿𝐿 ≅ 𝑇𝑇𝑟𝑟
𝑚𝑚𝑟𝑟 ≅ 0or
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12Prof. Sergio B. MendesSummer 2018
Example 5.4
𝑚𝑚𝑐𝑐 = 73 𝑘𝑘𝑘𝑘
𝑎𝑎𝑐𝑐 = ? ?
𝑚𝑚𝑟𝑟 = 940 𝑘𝑘𝑘𝑘
𝑡𝑡 = ? ?
∆𝑥𝑥 = 51 𝑚𝑚
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13Prof. Sergio B. MendesSummer 2018
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
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14Prof. Sergio B. MendesSummer 2018
Example 5.5
𝑣𝑣 = ? ?
𝑇𝑇 = ? ?
𝑚𝑚
𝜃𝜃
𝐿𝐿
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15Prof. Sergio B. MendesSummer 2018
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
=
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16Prof. Sergio B. MendesSummer 2018
Example 5.6
𝑣𝑣 = 90𝑘𝑘𝑚𝑚𝑒
= 25𝑚𝑚𝑠𝑠
𝜃𝜃 = ? ?
𝑟𝑟 = 350 𝑚𝑚
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17Prof. Sergio B. MendesSummer 2018
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°
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18Prof. Sergio B. MendesSummer 2018
𝑣𝑣 = ? ?
𝑟𝑟 = 6.3 𝑚𝑚
Example 5.7
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19Prof. Sergio B. MendesSummer 2018
𝑭𝑭𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑭𝑭𝒈𝒈 + 𝒏𝒏 = 𝑚𝑚 𝒂𝒂
𝑦𝑦: 𝑚𝑚𝑘𝑘 + 𝑛𝑛 = 𝑚𝑚 𝑎𝑎𝑦𝑦
𝑎𝑎𝑦𝑦 =𝑣𝑣2
𝑟𝑟
𝑣𝑣𝑚𝑚𝑚𝑚𝑛𝑛 = 𝑟𝑟 𝑘𝑘𝑣𝑣 = 𝑟𝑟 𝑘𝑘 +𝑛𝑛𝑚𝑚
𝑣𝑣𝑚𝑚𝑚𝑚𝑛𝑛 = 7.9𝑚𝑚𝑠𝑠
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20Prof. Sergio B. MendesSummer 2018
Friction Force
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21Prof. Sergio B. MendesSummer 2018
Walking
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22Prof. Sergio B. MendesSummer 2018
𝑓𝑓𝑠𝑠 ≤ 𝜇𝜇𝑠𝑠 𝑛𝑛
𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑛𝑛
𝜇𝜇𝑠𝑠 > 𝜇𝜇𝑘𝑘
Static and Kinetic Friction
Static
Kinetic
𝑓𝑓𝑠𝑠, 𝑚𝑚𝑚𝑚𝑥𝑥 = 𝜇𝜇𝑠𝑠 𝑛𝑛
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23Prof. Sergio B. MendesSummer 2018
Example 5.10
𝜇𝜇𝑠𝑠 = 0.46 𝜃𝜃 = ? ?
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24Prof. Sergio B. MendesSummer 2018
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°
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25Prof. Sergio B. MendesSummer 2018
Example 5.11
𝜇𝜇𝑘𝑘
𝑇𝑇 = ? ?
𝑚𝑚
𝜃𝜃
𝑣𝑣 = 𝑐𝑐𝑐𝑐𝑛𝑛𝑠𝑠𝑡𝑡𝑎𝑎𝑛𝑛𝑡𝑡
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26Prof. Sergio B. MendesSummer 2018
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
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27Prof. Sergio B. MendesSummer 2018
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
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28Prof. Sergio B. MendesSummer 2018
Drag Forces
Terminal Velocity
Depends on:
• Instantaneous velocity
• Area opposing the motion
Air Drag