Question

question 23
(problem reference m.2) two force vectors are shown in the figure. the magnitudes of the forces are a = 100 n and b = 125 n.
what is the y - component of the net force?
110 n
-110 n
225 n
-55.6 n
135 n

Explanation:

Step1: Find \( y \)-component of \( \vec{A} \)

The \( y \)-component of a vector \( \vec{V} \) with magnitude \( V \) and angle \( \theta \) (measured from the \( y \)-axis) is \( V_y = V\cos\theta \). For \( \vec{A} \), \( A = 100\ N \) and \( \theta_A=35.0^{\circ} \), so \( A_y=100\cos(35.0^{\circ})\).
Using a calculator, \( \cos(35.0^{\circ})\approx0.819 \), so \( A_y = 100\times0.819 = 81.9\ N \).

Step2: Find \( y \)-component of \( \vec{B} \)

The \( y \)-component of \( \vec{B} \) with magnitude \( B = 125\ N \) and angle \( \theta_B = 25.0^{\circ} \) (measured from the \( x \)-axis) is \( B_y=B\sin(25.0^{\circ})\).
Using a calculator, \( \sin(25.0^{\circ})\approx0.423 \), so \( B_y=125\times0.423 = 52.9\ N \).

Step3: Find the net \( y \)-component

The net \( y \)-component \( F_y=A_y + B_y\).
\( F_y=81.9+52.9=134.8\approx135\ N \)

Answer:

135 N