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What is the magnitude of the third force?

Two forces act on an object. One has a magnitude of 166 newtons and points at an angle of 60.0 degrees...?

  • Two forces act on an object. One has a magnitude of 166 newtons and points at an angle of 60.0 degrees above the +x axis. The second has a magnitude of 284 newtons and points at an angle of 30.0 degrees above the +x axis. The third force is applied and balances to zero the effects of the other two. What are the magnitude and direction of the third force? Specify the direction relative to the negative x axis. HELP...idk what im doing :(

  • Answer:

    You have two triangles of force, one at 60* and the other at 30*. To add these together we need the horizontal and vertical components of the vectors. For the vertical components Sin 60 = Vert / 166 and Sin 30 = Vert / 284 For the horizontal components Cos 60 = Hor / 166 and Cos 30 = Hor / 284 Vertical values of 143.76 and 142 Horizontal values of 83 and 245.95 So the resultant force is a vertical 285.76 and horizontal 328.95 Using Pythagoras the Hypotenuse value = sqrt. ( 285.76^2 + 328.95^2) Resultant force = 435.74N The angle this force acts at, Tan X = 285.76 / 328.95 Resultant force angle = 40.98 degrees So your balancing force will be 435.74N at (180 + 40.98) = 220.98 degrees. GL>

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You have two triangles of force, one at 60* and the other at 30*. To add these together we need the horizontal and vertical components of the vectors. For the vertical components Sin 60 = Vert / 166 and Sin 30 = Vert / 284 For the horizontal components Cos 60 = Hor / 166 and Cos 30 = Hor / 284 Vertical values of 143.76 and 142 Horizontal values of 83 and 245.95 So the resultant force is a vertical 285.76 and horizontal 328.95 Using Pythagoras the Hypotenuse value = sqrt. ( 285.76^2 + 328.95^2) Resultant force = 435.74N The angle this force acts at, Tan X = 285.76 / 328.95 Resultant force angle = 40.98 degrees So your balancing force will be 435.74N at (180 + 40.98) = 220.98 degrees. GL>

ALEX

For the best answers, search on this site https://shorturl.im/Dofvb Use vector arithmetic to add the first two. Get the x and y components of each and add those to get the x and y components of the sum. Now just reverse the polarity of those x and y components of the sum above to get the third force. To change those to magnitude and direction, use the following: R = √(Rx² + Ry²) θ = arctan (Ry/Rx) .

Kathleen

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