Question

examine the free body diagram shown above. what is the net force on the object depicted in this diagram? 55 n, 29° above the negative x - axis 70 n, 36° above the negative x - axis 61 n, 7° below the negative x - axis 24 n, 19° above the positive x - axis

Explanation:

Step1: Calculate the x - components of the forces

The x - component of a force \(F\) with an angle \(\theta\) (measured from the x - axis) is given by \(F_x = F\cos\theta\).
For the \(900N\) force: \(\theta = 135^{\circ}\) (since it is \(45^{\circ}\) above the negative x - axis), \(F_{1x}=900\cos135^{\circ}=900\times(-\frac{\sqrt{2}}{2})\approx - 636.4N\)
For the \(600N\) force: \(\theta = 15^{\circ}\), \(F_{2x}=600\cos15^{\circ}\approx600\times0.966 = 579.6N\)
The sum of the x - components \(F_x=F_{1x}+F_{2x}\approx-636.4 + 579.6=-56.8N\)

Step2: Calculate the y - components of the forces

The y - component of a force \(F\) with an angle \(\theta\) (measured from the x - axis) is given by \(F_y = F\sin\theta\).
For the \(900N\) force: \(F_{1y}=900\sin135^{\circ}=900\times\frac{\sqrt{2}}{2}\approx636.4N\)
For the \(600N\) force: \(F_{2y}=600\sin15^{\circ}\approx600\times0.259 = 155.4N\)
The \(750N\) force acts along the negative y - axis, so \(F_{3y}=-750N\)
The sum of the y - components \(F_y=F_{1y}+F_{2y}+F_{3y}\approx636.4 + 155.4-750=41.8N\)

Step3: Calculate the magnitude of the net force

The magnitude of the net force \(F=\sqrt{F_x^{2}+F_y^{2}}\)
\(F=\sqrt{(- 56.8)^{2}+41.8^{2}}=\sqrt{3226.24 + 1747.24}=\sqrt{4973.48}\approx70N\)

Step4: Calculate the direction of the net force

The direction \(\theta=\tan^{-1}(\frac{F_y}{|F_x|})\)
\(\theta=\tan^{-1}(\frac{41.8}{56.8})\approx36^{\circ}\) above the negative x - axis

Answer:

70 N, \(36^{\circ}\) above the negative \(x\) - axis