Can someone plz explain this more thoroughly. I need help with math and concepts..Q3 The given force we have is:700N at an angle 15° anticlockwise from -y axis.And the required resultant force is 500N directed along +y axis.Now, the x component of the force F should be equal to the x component of force 700N since there is noresultant force in the x direction.→700 cos 15° = F cos 0F cos 0 = 676.14 NStep2b)And the net force in the y axis should be equal to 500N.→F sin 0 – 700 sin 15° = 500→F sin 0 = 500 + 181.17→F sin 0 = 681.17NSo, the angle of force F should be:F sin 0681.17'F cos 0(676.14,(681.17)tan 0 =\676.14,>0 = 45.212°And the force F will be:F cos 45.212°= 676.14→F = 959.76NTherefore, The force F is 959.76N and the angle 0 is 45.212° 2) In Figure 1,force and its direction, measured counterclockwise from the positive x-axis.60° and F = 450 N, determine the magnitude of the resultantF,15°700 NFigure 13) If the magnitude of the resultant force is to be 500 N, directed along thepositive y-axis, determine the magnitude of force F and its direction O inFigure 1.

Answered: Image /qna-images/answer/400d5667-a395-4e50-8b78-4f80a9a18bfc.jpg

Answered: 3) If the magnitude of the resultant…

3) If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction O in Figure 1.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

Can someone plz explain this more thoroughly. I need help with math and concepts..Q3

The given force we have is:
700N at an angle 15° anticlockwise from -y axis.
And the required resultant force is 500N directed along +y axis.
Now, the x component of the force F should be equal to the x component of force 700N since there is no
resultant force in the x direction.
→700 cos 15° = F cos 0
F cos 0 = 676.14 N
Step2
b)
And the net force in the y axis should be equal to 500N.
→F sin 0 – 700 sin 15° = 500
→F sin 0 = 500 + 181.17
→F sin 0 = 681.17N
So, the angle of force F should be:
F sin 0
681.17'
F cos 0
(676.14,
(681.17)
tan 0 =
\676.14,
>0 = 45.212°
And the force F will be:
F cos 45.212°
= 676.14
→F = 959.76N
Therefore, The force F is 959.76N and the angle 0 is 45.212°

2) In Figure 1,
force and its direction, measured counterclockwise from the positive x-axis.
60° and F = 450 N, determine the magnitude of the resultant
F,
15°
700 N
Figure 1
3) If the magnitude of the resultant force is to be 500 N, directed along the
positive y-axis, determine the magnitude of force F and its direction O in
Figure 1.

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