Respuesta :
Explanation:
The net force is calculated as follows.
[tex]F_{net} = F_{1} + F_{2} + F_{3}[/tex]
Also, [tex]F_{net}[/tex] = ma
= [tex]2 \times 3[/tex]
= 6 N
= [tex]6 Cos 50 \hat{i} + 6 Sin 50 \hat{j}[/tex]
= [tex]3.86 \hat{i} + 4.6 \hat{j}[/tex]
[tex]F_{1} = -10 Cos (30)\hat{i} - 10 Sin (30)\hat{j}[/tex]
= [tex]-8.66\hat{i} - 5\hat{j}[/tex]
[tex]F_{2} = 20\hat{j}[/tex]
[tex]3.86 \hat{i} + 4.6 \hat{j}[/tex] = [tex](-8.66\hat{i} - 5\hat{j}) + 20\hat{j} + F_{3}[/tex]
[tex]F_{3} = 12.52\hat{i} - 10.4\hat{j}[/tex]
= 16.3 N
Thus, we can conclude that the third force [tex]F_{3}[/tex] in unit vector notation is 16.3 N and magnitude angle notation is [tex]39.7^{o}[/tex].
Answer:
The third force [tex]F_{3}[/tex] in unit vector notation and in magnitude angle notation is [tex]39.7[/tex]°
Explanation:
2 kg cookie tin acceleration [tex]= 3 m/s^2[/tex] towards N-E
Angle = [tex]50[/tex]°
[tex]F_{1} = 10 N[/tex], South-West 30° off West
[tex]F_{2} = 20 N[/tex] towards North
[tex]F_{3} = ?[/tex]
Step 1:
Total Force,
[tex]$\begin{aligned} F_{\text {het }} &=F_{1}+F_{2}+F_{3} . \\ F_{\text {net }} &=m a=2(3)=6 \mathrm{~N} \\ &=6 \cos 50 \hat{\imath}+6 \sin 50 \hat{\jmath} \\ &=3.86 \hat{\imath}+4.6 \hat{\jmath} \end{aligned}$[/tex]
[tex]$\begin{aligned} F_{1} &=-10 \cos 30 \hat{\imath}-10 \sin 30 \hat{\jmath} \\ &=-8.66 \hat{1}-5 \hat{\jmath}^{.} \\ F_{2} &=20 \hat{\jmath} . \end{aligned}$[/tex]
Step 2:
Calculate the third force :
[tex]$3 \cdot 86 \hat{\imath}+4 \cdot 6 \hat{\jmath}=(-8.66 \hat{\imath}-5 \hat{j})+20 \hat{\jmath}+F_{3}$[/tex]
[tex]$F_{3}=12.52 \hat{-10.4 j}$[/tex]
[tex]$F_{3}=16.3 \mathrm{~N}$[/tex]
Direction[tex]= $39.7^{\circ}$[/tex] South of east.
Thus we conclude that the third force in unit vector notation and in magnitude angle notation is [tex]$39.7^{\circ}$[/tex].
To learn more about Force, refer:
- https://brainly.com/question/18038995
- https://brainly.com/question/18775450
