Three forces in the x-y plane act on a 3.90 kg mass: 11.70 N...

Three forces in the x-y plane act on a 3.90 kg mass: 11.70 N directed at 41°, 11.20 N directed at 160°, and 10.40 N directed at 227°. All angles are measured from the positive x-axis, with positive angles in the Counter-Clockwise direction. Calculate the magnitude of the acceleration. 2.465 m/s^2 Calculate the direction of the resultant force using the same sign convention as above (in degrees). 200

Transcript

00:01 In this question we have given three forces in xy plane which act on a 3 .9 kilogram mass then we have to find the magnitude of the acceleration of this mass so given that m is 3 .9 kilogram so if we light the first force vector that is f1 vector is we have given that these are the three mass forces which act on this mass one is at angle of 41 degree another one is at an angle of 160 degree that is in second quadrant and the third one is in third quadrant so these are the three forces this angle is 41 degree from positive xx is this is 160 degree and this was is 227 degree this value of force is 11 .7 newton let this is f1 f2 is the magnitude this is our f2 and the magnitude is this is this is 11 .2 and this was f3 whose magnitude is 10 .4 and this is the mass m so if we write the vector form so we can say that f1 vector is this is 11 .7 that is the magnitude into cos 41 i cap plus sine 41 degree j cap this is the in newton this is f1 similarly we can say that f2 vector can be written as this is magnitude is 11 .2 into unit vector dcose 60 degree i cap plus sine 160 degree j cap and f3 vector is 10 .9 this is 10 .4 cos 2 to 7 degree i cap plus sign to 27 degree j cap this over r in newton so if we calculate this is 10 .4 so the resultant force is given as that is f vector is f1 vector plus f2 vector plus f3 vector so we have to add all the three vectors we can say this will comes out 8 .96 minus 10 .5 minus 7 .099 i cap plus 7 .52 plus 3 .83 minus 7 .61 j cap and that will come out minus 8 .81 i cap plus 3 .84 j cap.

03:21 This is the resultant force vectors.

03:27 The magnitude of this force vector is given by that is under root of minus 8 .81 square plus 3 .84 square and that will comes out 9 .6 so this is the net force.

03:53 So if we calculate the acceleration, so acceleration is given by that is a is equal to the net force upon mass.

04:05 So this is 9 .61 upon 3 .9 and that will comes out 2 .465 meter per second square.

04:14 So this is the acceleration...

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