Question

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 net force acting on the object. vectors ( f_1 ) and ( f_2 ) are shown in the diagram (not necessarily drawn to scale). their magnitudes and angles are:
( f_1 = 16.3 mathrm{n}, \theta_1 = 17^{circ} )
( f_2 = 17.1 mathrm{n}, \theta_2 = 61^{circ} )
find the ( y )-component ( r_y ) of the resultant vector in newtons. express your answer to two decimal places, and be sure it has the correct sign.
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 them so that they are tip - to - tail.

Explanation:

Step1: Find the y - component of \(F_1\)

The y - component of a vector \(F\) with magnitude \(F\) and angle \(\theta\) (measured from the x - axis) is given by \(F_y = F\sin\theta\). For \(F_1 = 16.3N\) and \(\theta_1=17^{\circ}\), \(F_{1y}=16.3\sin(17^{\circ})\).
Using a calculator, \(\sin(17^{\circ})\approx0.292\), so \(F_{1y}=16.3\times0.292 = 4.76N\).

Step2: Find the y - component of \(F_2\)

For \(F_2 = 17.1N\) and \(\theta_2 = 61^{\circ}\), since the vector \(F_2\) is below the x - axis, its y - component is negative. \(F_{2y}=- 17.1\sin(61^{\circ})\).
Using a calculator, \(\sin(61^{\circ})\approx0.875\), so \(F_{2y}=-17.1\times0.875=-15.07N\).

Step3: Find the y - component of the resultant vector \(R_y\)

By the principle of vector addition, \(R_y=F_{1y}+F_{2y}\).
Substitute \(F_{1y} = 4.76N\) and \(F_{2y}=-15.07N\) into the formula: \(R_y=4.76+( - 15.07)=-10.31N\).

Answer:

\(-10.31\)