Answered: Based on the diagram attached,…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter26: Relativity

Section: Chapter Questions

Problem 33P: A chain of nuclear reactions in the Suns core converts four protons into a helium nucleus. (a) What...

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Question

Based on the diagram attached, determine and state the magnitude and direction of the net force
produced relative to the horizontal, when the three forces shown below
act at a single point.

Please insert formulas

Expert Solution

Step 1

Scenario - Forces are applied to different directions. Their resultant will be in accordance with vector addition law.

Given,

$F_{1} = 10 N F_{2} = 15 N F_{3} = 20 N θ_{1} = 35 ° θ_{2} = 60 ° θ_{3} = 90 °$

Step 2

Formula Used:

Parallelogram Law for vector addition $\to {|\vec{C}|}^{2} = {|\vec{A}|}^{2} + {|\vec{B}|}^{2} + 2 \vec{|A|} |\vec{B}| \cos θ$

and angle between them is $α = \tan^{- 1} (\frac{|\vec{B}| \sin θ}{|\vec{B}| + |\vec{A}| \cos θ})$

Step 3

Finding resultant of $F_{1}$ and $F_{2}$ as :

Step 4

$\begin{array}{rcl} {|\vec{F}|}^{2} & = & {|{\vec{F}}_{1}|}^{2} + {|\vec{F_{2}}|}^{2} + 2 |{\vec{F}}_{1}| |{\vec{F}}_{2}| \cos θ \\ = & \sqrt{10^{2} + 15^{2} + 2 \times 10 \times 15 \times \cos (180 - 60 - 35) °} \\ = & \sqrt{100 + 225 + 300 \times \cos 85} \\ = & \sqrt{325 + 300 \times 0.087} \\ = & \sqrt{325 + 26.1} \\ = & \sqrt{351.1} \\ \Rightarrow & |\vec{F}| = 18.73 N \end{array}$

$α = \tan^{- 1} (\frac{|{\vec{F}}_{2}| \sin θ}{|{\vec{F}}_{1}| + |\vec{F_{2}}| \cos θ}) \Rightarrow α = \tan^{- 1} (\frac{15 \times 0.996}{10 + 15 \times 0.0871}) \Rightarrow α = \tan^{- 1} (\frac{14.94}{10 + 1.3065}) = \tan^{- 1} (\frac{14.94}{11.307}) \Rightarrow α = t {an}^{- 1} (1.321) = 52.87 °$

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