Question

the free body diagram above shows the forces acting on an 84.1 kg object. what is the acceleration vector that results from these forces? 1.51 m/s², 18.9° east of north 1.88 m/s², 26.6° east of north 3.92 m/s², 23.3° west of north 2.05 m/s², 9.45° east of north

Explanation:

Step1: Resolve forces into components

• East - West direction:
• The force of \(120\ N\) has a west - ward component \(F_{x1}=- 120\cos30^{\circ}\ N\approx - 103.92\ N\), and the force of \(145\ N\) has an east - ward component \(F_{x2}=145\ N\).
• The net force in the \(x\) - direction \(F_x=F_{x2}+F_{x1}=145 - 103.92=41.08\ N\)
• North - South direction:
• The force of \(120\ N\) has a north - ward component \(F_{y1}=120\sin30^{\circ}\ N = 60\ N\), and the force of \(60\ N\) has a north - ward component \(F_{y2}=60\ N\).
• The net force in the \(y\) - direction \(F_y=F_{y1}+F_{y2}=60 + 60=120\ N\)

Step2: Calculate the magnitude of the net force

Using the Pythagorean theorem \(F=\sqrt{F_x^{2}+F_y^{2}}\)
\(F=\sqrt{41.08^{2}+120^{2}}=\sqrt{1687.5664 + 14400}=\sqrt{16087.5664}\approx126.84\ N\)

Step3: Calculate the magnitude of acceleration

According to Newton's second law \(F = ma\), where \(m = 84.1\ kg\)
\(a=\frac{F}{m}=\frac{126.84}{84.1}\approx1.51\ m/s^{2}\)

Step4: Calculate the direction of the net force (and thus the acceleration, since \(a=\frac{F}{m}\) and \(m>0\))

Using the formula \(\theta=\tan^{- 1}(\frac{F_x}{F_y})\)
\(\theta=\tan^{-1}(\frac{41.08}{120})\approx18.9^{\circ}\) east of north

Answer:

1.51 m/s², 18.9° east of north