Question

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?

Answer 1

Establishing a reference system:

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

Expressing each Force as a vector

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

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

Components of the resulting vector $\vec{R}$

• Horizontal component

$cos(27^{0})=\frac{ F_{R}}{F_{Rx} }\\ F_{Rx}=F_{R}*cos(27^{0})\\ F_{Rx}=9lb*cos(27^{0})\\F_{Rx}=8lb$

• Vertical component.

$sin(27^{0})=\frac{ F}{F_{Ry }}\\ F_{Ry}=R*sin(27^{0})\\ F_{Ry}=9lb*sin(27^{0})\\F_{Ry}=4lb$

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

The sum of the vectors is $\vec{F_{R}}=\vec{F_{A}}+\vec{F_{B}}+\vec{F_{C}}$

The force Fc is

: $\vec{F_{R}}=\vec{F_{A}}+\vec{F_{B}}+\vec{F_{C}}\\ \vec{F_{C}}=\vec{F_{R}}-\vec{F_{A}}-\vec{F_{B}}$

Evaluating numerically

$\vec{F_{C}}=(8\hat{i}+4\hat{j})lb-4(\hat{j})lb-14(\hat{i})lb\\ \vec{F_{C}}=(8\hat{i}-14\hat{i}+4\hat{j}-4\hat{j})lb \\ \vec{F_{C}}=-6\hat{i}lb$

The magnitude is

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

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