David is pulling on one limb with a force of 20 N at an angle of30 o North of Ea

Question

David is pulling on one limb with a force of 20 N at an angle of30o North of East. Mike is pulling with 30 N at40o North of West. Bill is pulling with 40 NNorth. Luke is pulling with a force of 50 N West. Lindapulled in such a way as to cancel the net force and create no netforce.

Graphically solve for net (resultant) force of these 4forces. What is the x and y component for the four forces andthe magnitude and direction of the net (or resultant) force? Findthe magnitude and direction of the force by Linda and what is thatforce called?

thank you

Explanation / Answer

X component of the resultantforce=Fx=20cos30+0-30cos40-50=17.32+0-22.98-50=-55.66N Y component of the resultantforce=Fy=20sin30+40-30sin40-0=10+40-0=19.28N net force magnitude=55.66*55.66+19.28*19.28=58.9N direction=tan=19.28/58.9=>=18.12degree the resultant force is making an angle of 18.12degree with X axisin clock wise direction as the resultant of all these forces [including force applied bylinda]is zero the force applied by linda is exactly equal to 58.9N at an angle of 18.12 degree opposite the the net force of allthe four forces.

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