Newton's Second Law — Question 8

Section 5-5 · Physics pages 706–707

Question 8 A desired acceleration determines the required control force Use ∑⃗F = m⃗a for constant mass in an inertial frame. A 5 kg body must have acceleration (2, −1) m/s2. An unavoidable environmental force is (−3, 4) N. A controllable force is the only other interaction. 5 kg⃗ FE = (−3, 4) N⃗ a = (2, −1) m/s2 Control force: |⃗FC| ≤15 N Schematic; labels specify the data. Arrow lengths are not a common scale. (a) Find the required total force. (b) Find the control force and its magnitude. (c) Can a controller limited to 15 N achieve the target?
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Question 8 - Solution A desired acceleration determines the required control force 5 kg⃗ FE = (−3, 4) N⃗ a = (2, −1) m/s2 Control force: |⃗FC| ≤15 N (a) Translate the target into a resultant.⃗ Freq = m⃗a = 5(2, −1) = (10, −5) N⃗ Frequired = 5(2, −1) = (10, −5) N. (b) Cancel the environment and supply the remainder.⃗ FC =⃗Freq −⃗FE = (13, −9) N , FC = √ 132 + 92 = 15.811 N⃗ FC = (10, −5) −(−3, 4) = (13, −9) N, magnitude √ 250 = 5 √ 10 ≈15.811 N. (c) Compare the required magnitude with the bound. F 2 C = 250 N2 > 152 = 225 N2 No. The unique required vector exceeds the 15 N limit. Checking only the 13 N x component would miss the simultaneous vertical demand.

Question 8

A desired acceleration determines the required control force

Use ∑F→=ma→\sum\vec F=m\vec a for constant mass in an inertial frame. A 5 kg body must have acceleration (2,−1)(2,-1) m/s2^2. An unavoidable environmental force is (−3,4)(-3,4) N. A controllable force is the only other interaction.

Diagram shown in the original SVG worksheet.

  1. Find the required total force.

  2. Find the control force and its magnitude.

  3. Can a controller limited to 15 N achieve the target?

Question 8 - Solution

A desired acceleration determines the required control force

Diagram shown in the original SVG worksheet.

(a) Translate the target into a resultant.

F→req=ma→=5(2,−1)=(10,−5)N\vec F_{\mathrm{req}}=m\vec a=5(2,-1)=(10,-5)\,\mathrm N F→required=5(2,−1)=(10,−5)\vec F_{\mathrm{required}}=5(2,-1)=(10,-5) N.

(b) Cancel the environment and supply the remainder.

F→C=F→req−F→E=(13,−9)N,FC=132+92=15.811N\boxed{\vec F_C=\vec F_{\mathrm{req}}-\vec F_E=(13,-9)\,\mathrm N},\quad F_C=\sqrt{13^2+9^2}=15.811\,\mathrm N F→C=(10,−5)−(−3,4)=(13,−9)\vec F_C=(10,-5)-(-3,4)=(13,-9) N, magnitude 250=510≈15.811\sqrt{250}=5\sqrt{10}\approx15.811 N.

(c) Compare the required magnitude with the bound.

FC2=250N2>152=225N2F_C^2=250\,\mathrm{N^2}>15^2=225\,\mathrm{N^2} No. The unique required vector exceeds the 15 N limit. Checking only the 13 N x component would miss the simultaneous vertical demand.