Question

question 24
2 points
(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 magnitude of the net force?
146 n
25 n
191 n
223 n
225 n

Explanation:

Step1: Calculate the x - components of the vectors

For vector \(\vec{A}\), \(A_x=-A\sin35^{\circ}\) (negative because it is in the negative x - direction). Given \(A = 100\ N\), so \(A_x=- 100\sin35^{\circ}\approx - 100\times0.574=-57.4\ N\).
For vector \(\vec{B}\), \(B_x = B\cos25^{\circ}\). Given \(B = 125\ N\), so \(B_x=125\cos25^{\circ}\approx125\times0.906 = 113.25\ N\).
The x - component of the net force \(F_x=A_x + B_x=-57.4+113.25 = 55.85\ N\).

Step2: Calculate the y - components of the vectors

For vector \(\vec{A}\), \(A_y=A\cos35^{\circ}\). Given \(A = 100\ N\), so \(A_y=100\cos35^{\circ}\approx100\times0.819 = 81.9\ N\).
For vector \(\vec{B}\), \(B_y = B\sin25^{\circ}\). Given \(B = 125\ N\), so \(B_y=125\sin25^{\circ}\approx125\times0.423=52.875\ N\).
The y - component of the net force \(F_y=A_y + B_y=81.9+52.875 = 134.775\ N\).

Step3: Calculate the magnitude of the net force

Using the Pythagorean theorem \(F=\sqrt{F_x^{2}+F_y^{2}}\).
\(F=\sqrt{(55.85)^{2}+(134.775)^{2}}=\sqrt{3119.2225 + 18167.3006}=\sqrt{21286.5231}\approx146\ N\).

Answer:

146 N