A 3.00 kg mass is acted upon by four forces in the horizontal (xy) plane, where

Question

A 3.00 kg mass is acted upon by four forces in the horizontal (xy) plane, where F1 = 36.0 N (45 deg North of West), F2 = 51.0 N (30 deg North of East), F3 = 24.0 N (50 deg South of West), and F4 = 45.0 N (20 deg South of East). Find the acceleration of the mass and give both magnitude and direction above the positive x-axis.

What are the steps for finding this? I know the steps with just two forces, but I keep getting the answer wrong and can't tell what I'm doing wrong. I would appreciate an explanation of how to set this problem up and solve it!

Explanation / Answer

Given mass ,m = 3.0kg F1 = -36 cos45 i + 36 sin45 j= -25.45 i+25.45 j F2 = 51 cos 30 i + 51 sin30 j= 44.16 i+25.5 j F3 = -24 cos50 i - 24 sin 50j =-15.42 i-18.38 j F4 = 45 cos20 i - 45sin 20 j =42.28 i-15.39 j Fx = -25.45+44.16-15.42+42.28 =45.57 Fy =25.45+25.5-18.38-15.39 =17.18 F =45.57 i+17.18 j Since F =ma a =F/m = (45.57 i+17.18 j)/3.0kg =15.19 i+5.72 j Magnitude of a =v(15.19)^2+(5.72)^2 =16.23 m/s^2 Direction ? = tan^-1(5.72/15.19) ? =20.63^0

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