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Solved: Angle for Minimum Force. A box with weight w is
Chapter 5, Problem 121CP(choose chapter or problem)
Angle for Minimum Force. A box with weight w is pulled at constant speed along a level floor by a force \(\vec{F}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{k}\).
(a) In terms of \(\theta, \mu_{k}\), and w, calculate 𝐹.
(b) For w = 400 N and \(\mu_{\mathrm{k}}=0.25\), calculate 𝐹 for \(\theta\) ranging from
\(0^{\circ} \text { to } 90^{\circ}\) in increments of \(10^{\circ}\). Graph 𝐹 versus \(\theta\).
(c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of 𝐹, required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here 𝐹 is a function of θ. ) For the special case of w = 400 N and \(\mu_{\mathrm{k}}=0.25\), evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part (b).
Equation Transcription:
0° to 90°
10°
Text Transcription:
\vec{F}
\theta
\mu_{k}
\theta, \mu_{k}
\mu_{\{k}}=0.25
\theta
0^° to 90^°
10^°
\(\theta\)
\theta
\mu_{\{k}}=0.25
\theta
Questions & Answers
QUESTION:
Angle for Minimum Force. A box with weight w is pulled at constant speed along a level floor by a force \(\vec{F}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{k}\).
(a) In terms of \(\theta, \mu_{k}\), and w, calculate 𝐹.
(b) For w = 400 N and \(\mu_{\mathrm{k}}=0.25\), calculate 𝐹 for \(\theta\) ranging from
\(0^{\circ} \text { to } 90^{\circ}\) in increments of \(10^{\circ}\). Graph 𝐹 versus \(\theta\).
(c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of 𝐹, required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here 𝐹 is a function of θ. ) For the special case of w = 400 N and \(\mu_{\mathrm{k}}=0.25\), evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part (b).
Equation Transcription:
0° to 90°
10°
Text Transcription:
\vec{F}
\theta
\mu_{k}
\theta, \mu_{k}
\mu_{\{k}}=0.25
\theta
0^° to 90^°
10^°
\(\theta\)
\theta
\mu_{\{k}}=0.25
\theta
ANSWER:Solution 121CP
Step 1
This is the representation of a weight, w moves in a an inclined plane.