53. (a) Using Eq. 22-28, we find GF=××+×−=−−−8 00 108 00 106000 2400 048055.±.±.±.±C3.0010NCiCNCjNiNj.3chchbgbgbgTherefore, the force has magnitude equal to ()()220.240N0.0480N0.245N.xyFFF=+=+−=(b) The angle the force FGmakes with the +xaxis is 110.0480Ntantan11.30.240NyxFFθ⎛⎞−===−°⎜⎟⎝⎠measured counterclockwise from the +xaxis. (c) With m= 0.0100 kg, the (x, y) coordinates at t= 3.00 s can be found by combining
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