A force F of magnitude 5.10 units acts on an object at the origin in a direction 0 = 38.0° above the positive x-axis. (See the figure below.) A second force F2 ofmagnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant forceF12unitsmagnitudeocounterclockwise from the +x-axisdirection

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Asked Oct 4, 2019

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A force F of magnitude 5.10 units acts on an object at the origin in a direction 0 = 38.0° above the positive x-axis. (See the figure below.) A second force F2 of
magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force
F
1
2
units
magnitude
ocounterclockwise from the +x-axis
direction
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A force F of magnitude 5.10 units acts on an object at the origin in a direction 0 = 38.0° above the positive x-axis. (See the figure below.) A second force F2 of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force F 1 2 units magnitude ocounterclockwise from the +x-axis direction

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Expert Answer

Step 1

The resultant force of F1 and F2 is resolved into F1cosθ along the horizontal direction , and ...

The resultant x component of force F and F, is given by
F F cose
=(5.10N) cos38.0°
=4.01N
The resultant y component of force F and F,is given by
F
F sin 0F
= (5.10N)sin 38.0° 5.00 N
=8.13N
The magnitude of the force is given by
(4.01N) + (8.13N)
F=A
=9.06N
The direction of the force is given by
8.13 N
0=tan
4.01 N
= 63.740
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The resultant x component of force F and F, is given by F F cose =(5.10N) cos38.0° =4.01N The resultant y component of force F and F,is given by F F sin 0F = (5.10N)sin 38.0° 5.00 N =8.13N The magnitude of the force is given by (4.01N) + (8.13N) F=A =9.06N The direction of the force is given by 8.13 N 0=tan 4.01 N = 63.740

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Newtons Laws of Motion

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