Answer to Question #94713 in Mechanics | Relativity for Haley T

Question #94713
Three forces act together on an object: Force A of 4.0 lb due South, Force B of 14lb due East, and Force C to be determined. The result of these three forces is a net force R of 9.0 lb in a direction 27 degrees north of east. What is the magnitude and direction of Force C?
1
Expert's answer
2019-09-18T09:49:46-0400

Establishing a reference system:

  • North Direction: + y
  • South Direction -j
  • East Direction + i
  • West Direction -i

Expressing each Force as a vector


FA=4(j^)lbFB=14(i^)lb\vec{F_{A}}=-4(\hat{j})lb\\ \vec{F_{B}}=14(\hat{i})lb

FC=??\vec{F_{C}}=??

Components of the resulting vector R\vec{R}


  • Horizontal component

cos(270)=FRFRxFRx=FRcos(270)FRx=9lbcos(270)FRx=8lbcos(27^{0})=\frac{ F_{R}}{F_{Rx} }\\ F_{Rx}=F_{R}*cos(27^{0})\\ F_{Rx}=9lb*cos(27^{0})\\F_{Rx}=8lb

  • Vertical component.

sin(270)=FFRyFRy=Rsin(270)FRy=9lbsin(270)FRy=4lbsin(27^{0})=\frac{ F}{F_{Ry }}\\ F_{Ry}=R*sin(27^{0})\\ F_{Ry}=9lb*sin(27^{0})\\F_{Ry}=4lb


Finally FR=(8i^+4j^)lb\vec{F_R}=(8\hat{i}+4\hat{j})lb


The sum of the vectors is FR=FA+FB+FC\vec{F_{R}}=\vec{F_{A}}+\vec{F_{B}}+\vec{F_{C}}


The force Fc is

: FR=FA+FB+FCFC=FRFAFB\vec{F_{R}}=\vec{F_{A}}+\vec{F_{B}}+\vec{F_{C}}\\ \vec{F_{C}}=\vec{F_{R}}-\vec{F_{A}}-\vec{F_{B}}


Evaluating numerically


FC=(8i^+4j^)lb4(j^)lb14(i^)lbFC=(8i^14i^+4j^4j^)lbFC=6i^lb\vec{F_{C}}=(8\hat{i}+4\hat{j})lb-4(\hat{j})lb-14(\hat{i})lb\\ \vec{F_{C}}=(8\hat{i}-14\hat{i}+4\hat{j}-4\hat{j})lb \\ \vec{F_{C}}=-6\hat{i}lb


The magnitude is

FC=6ilbFC=6lb\vec{F_{C}}=|-6i|lb\\\vec{F_{C}}=6lb


The direction is West (-i)

The direction is determined by the unit vector of the force Fc expressed as a vector, which indicates -i that is, West





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