What is the magnitude of the total force on the ball?
5 boys are pushing on a snowball, and each is pushing with a force of 10.0N. however, each boy is pushing in a different direction. they are pushing north, northeast, east, southeast, and south. (Each boy is pushing at an angle of 45.0 degrees relative to his neighbor.)
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The total force on the ball is the sum of every individual force vector pushing on the ball.(note that we're excluding gravity, which would point down into the screen).
Note that in the y direction, all the forces will cancel. So the net force vector will be pointing in the x direction. We have 10 N + 2*10N*cos45 = 10N(1 + 2*sqrt(2)/2) = 24.14 N in the x direction.
The magnitude then, is 24.14, since the force in the y direction is zero.
Let A ,B C ,D and E be the persons who push the ball north , north east, east, south east and sothern directions. Resolving the force of each in North and eastern directions , we get:
A; 10 cos 0 = 10N towards north and 10 sin0 =0N east
B:10 cos 45=10/sqrt2Nnorth and 10 sin 45 =10/sqrt2N east
C: 10 cos90 =0N north and 10 sin90 = 10N east
D: 10 cos (-135)= -10/sqrt2 N and 10 sin135 = 10/sqrt2
E: 10 cos 180 = -10 N north and 10 sin 180 =0 east
Therefore: thesum of north components=(10+10/srt2+0-10/sqrt2-10)= 0 and the total of east components =2(10/sqrt2)+10 = 10(2/sqrt2 +10=10(sqrt2+1) N = 24.1421N approximately. Therefore the resultant of 0 north and 24.1421 N towards east is 24.1421 N and in eastern direction.
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