Question #232929

A=2000 lb, north

B=4000 lb, 60 degrees north of west

C=5000 lb, 50 degrees west of south

D= 1000 lb, south

E= 3000 lb, 60 degrees of south

Find the resultant displacement using component method


1
Expert's answer
2021-09-04T07:35:27-0400

Let's first find xx- and yy-components of resultant displacement:

Rx=2000 lbcos90+4000 lbcos150+5000 lbcos230+1000 lbcos270+3000 lbcos330=4080 lb,R_x=2000\ lb\cdot cos90^{\circ}+4000\ lb\cdot cos150^{\circ}+5000\ lb\cdot cos230^{\circ}+1000\ lb\cdot cos270^{\circ}+3000\ lb\cdot cos330^{\circ}=-4080\ lb,

Ry=2000 lbsin90+4000 lbsin150+5000 lbsin230+1000 lbsin270+3000 lbsin330=2330 lb.R_y=2000\ lb\cdot sin90^{\circ}+4000\ lb\cdot sin150^{\circ}+5000\ lb\cdot sin230^{\circ}+1000\ lb\cdot sin270^{\circ}+3000\ lb\cdot sin330^{\circ}=-2330\ lb.

We can find the magnitude of the resultant displacement from the Pythagorean theorem:


R=Rx2+Ry2=(4080 lb)2+(2330 lb)2=4698.5 lb.R=\sqrt{R_x^2+R_y^2}=\sqrt{(-4080\ lb)^2+(-2330\ lb)^2}=4698.5\ lb.

We can find the direction of the resultant displacement from the geometry:


θ=sin1(RyR),\theta=sin^{-1}(\dfrac{R_y}{R}),θ=sin1(2330 lb4698.5 lb)=30 W of S.\theta=sin^{-1}(\dfrac{-2330\ lb}{4698.5\ lb})=30^{\circ}\ W\ of\ S.

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