Question

formula 12 points
vectors ( f_1 ) and ( f_2 ) represent forces acting on an object. adding ( f_1 ) and ( f_2 ) gives us the resultant vector ( r ), which represents the ne
vectors ( f_1 ) and ( f_2 ) are shown in the diagram (not necessarily drawn to scale). their magnitudes and angles are:
( f_1 = 10.0 mathrm{~n}, \theta_1 = 47^{circ} )
( f_2 = 7.9 mathrm{~n}, \theta_2 = 37^{circ} )
find the magnitude of the resultant vector ( r ) in newtons. express your answer to two decimal places.

note: you are encouraged to add to the drawing here, or to create your own. remember, when adding vectors, it is often helpful to re - draw th
answer

Explanation:

Step1: Calculate \(x\) - components of \(F_1\) and \(F_2\)

The \(x\) - component of a vector \(F\) with magnitude \(|F|\) and angle \(\theta\) (measured from the positive \(x\) - axis) is \(F_x = |F|\cos\theta\).
For \(F_1\): \(F_{1x}=10.0\cos(47^{\circ})\). Using a calculator, \(\cos(47^{\circ})\approx0.682\), so \(F_{1x}=10.0\times0.682 = 6.82\space N\)
For \(F_2\): \(F_{2x}=- 7.9\cos(37^{\circ})\). Since \(F_2\) is in the second - quadrant. \(\cos(37^{\circ})\approx0.799\), so \(F_{2x}=-7.9\times0.799=-6.3121\space N\)
The \(x\) - component of the resultant vector \(R_x=F_{1x}+F_{2x}=6.82-6.3121 = 0.5079\space N\)

Step2: Calculate \(y\) - components of \(F_1\) and \(F_2\)

The \(y\) - component of a vector \(F\) with magnitude \(|F|\) and angle \(\theta\) (measured from the positive \(x\) - axis) is \(F_y = |F|\sin\theta\).
For \(F_1\): \(F_{1y}=10.0\sin(47^{\circ})\). Using a calculator, \(\sin(47^{\circ})\approx0.731\), so \(F_{1y}=10.0\times0.731 = 7.31\space N\)
For \(F_2\): \(F_{2y}=7.9\sin(37^{\circ})\). \(\sin(37^{\circ})\approx0.602\), so \(F_{2y}=7.9\times0.602 = 4.7558\space N\)
The \(y\) - component of the resultant vector \(R_y=F_{1y}+F_{2y}=7.31 + 4.7558=12.0658\space N\)

Step3: Calculate the magnitude of the resultant vector \(R\)

The magnitude of a vector \(R=(R_x,R_y)\) is given by \(|R|=\sqrt{R_x^{2}+R_y^{2}}\)
Substitute \(R_x = 0.5079\) and \(R_y = 12.0658\) into the formula:
\(|R|=\sqrt{(0.5079)^{2}+(12.0658)^{2}}=\sqrt{0.258 + 145.589}=\sqrt{145.847}\approx12.08\space N\)

Answer:

\(12.08\)