Question

The two forces F1 and F2 shown in the figure(looking down) act on a 22.0 kg object on a frictionless tabletop.

If F1 = 10.6 N and F2= 16.2 N, find the A) angle of the net force on the object for (a). B) absolute value of the net force on the object for (b). C)angle of the net force on the object for (b). D) acceleration of the object for (b).

Answer 1

Given

Mass of object \(m=22.0 \mathrm{~kg}\)

$$ \begin{array}{l} F_{1}=10.6 \mathrm{~N} \\ \mathrm{~F}_{2}=16.2 \mathrm{~N} \end{array} $$

From figure (a),

The net force is

$$ \begin{aligned} F_{a} &=-F_{1} \hat{i}-F_{2} \hat{j} \\ &=-(10.6 \mathrm{~N}) \hat{i}-(16.2 \mathrm{~N}) \hat{j} \end{aligned} $$

From figure \((b)\)

$$ \begin{aligned} F_{b} &=\left(F_{1} \cos 30^{\circ}\right) \hat{i}+\left[F_{2}-F_{1} \sin 30^{\circ}\right] \hat{j} \\ &=\left((10.6 \mathrm{~N}) \cos 30^{\circ}\right) \hat{i}+\left[(16.2 \mathrm{~N})-(10.6 \mathrm{~N}) \sin 30^{\circ}\right] \hat{j} \\ &=(9.179869 \mathrm{~N}) \hat{i}+(10.9 \mathrm{~N}) \hat{j} \end{aligned} $$

A)

The angle of the net force on the object for \((a)\)

$$ \begin{aligned} \phi_{a} &=\tan ^{-1}\left[\frac{-(16.2 \mathrm{~N})}{-(10.6 \mathrm{~N})}\right] \\ &=180^{\circ}+\tan ^{-1}\left[\frac{(16.2 \mathrm{~N})}{[(10.6 \mathrm{~N})}\right] \\ &=236.8^{\circ} \end{aligned} $$

B) The absolute value of the net force on the object for (b)

$$ \begin{aligned} \left|F_{b}\right| &=\sqrt{(9.179869 \mathrm{~N})^{2}+(10.9 \mathrm{~N})^{2}} \\ &=14.25 \mathrm{~N} \end{aligned} $$

C)

The angle of the net force on the object for (b)

$$ \begin{aligned} \phi_{6} &=\tan ^{-1}\left[\frac{10.9 \mathrm{~N}}{9.179869 \mathrm{~N}}\right] \\ &=49.89^{\circ} \end{aligned} $$

D)

Apply Newton's second law,

$$ F_{b}=m a $$

The acceleration object for (b)is

$$ \begin{aligned} a &=\frac{F_{b}}{m} \\ &=\frac{14.25 \mathrm{~N}}{22.0 \mathrm{~kg}} \\ &=0.65 \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$