CH 2: Force Vectors Assignment Week 1 2-33. If F1 = 600 N...

CH 2: Force Vectors Assignment Week 1 2-33. If F1 = 600 N and ? = 30°, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis.

Transcript

00:01 Here we have three forces acting on the eye bolt and we have to find out the resulted force magnitude and direction.

00:12 So the first force is f1, it has value of 600 newton and angle 5 is equal to 30 degree with positive x -axis.

00:25 Second force is f2 equal to 500 newton and it has angle 60 degree in clock wide direction with positive x -axis now third force is f3 equal to 450 newton and it will have angle theta with negative x -axis in counterclockwise direction now this angle theta is equal to theta here so from this we can find sine theta equal to 4 upon 5 in this triangle 5 is hypertenuous 4 is perpendicular and 3 is base so sine theta equal to 4 by 5 and cos theta is equal to 3 upon 5 now we have to find the result so we will add the 3 forces so net force is f net equal to f1 plus f2 plus f3 we'll write it in component form so magnitude is 600 newton and the x and y components will be cost 5 sorry cost 30 degree i plus sign 30 degree j cap 500 newton for f2 cost minus 60 degree because it is clockwise counterclockwise positive so clockwise direction is negative plus sign minus 60 degree j cap now the third one is f3 450 newton it will have x component negative toward negative x -axis so this is minus cost theta i now cost theta is 3 by 5 so it will be 3 by 5 i similarly the y cup rate would be also negative minus sine theta j so minus sine theta is equal to 4 .5 j cap.

03:09 Now we use the calculator.

03:13 So if 1 is 519 .6i plus 300 jcap.

03:23 F2 is 250 i minus 433 .0 .03 .013 .13.

03:39 And f3 will be minus 270 i minus 360 j.

03:58 So these are the three forces.

04:03 Now we will add them and we get 499 .6 to i minus 493 .013 j cap so these are the two x and y components so magnitude will be f net x square plus f net y square x square x square component of net force is 499 .62 y component is minus 493 .013 square so this gives us 701 .9 .9 newton this is the magnitude of the net force or resultant force...

Question

CH 2: Force Vectors Assignment Week 1 2-33. If F1 = 600 N and ? = 30°, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis.

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