A 2 kg cookie tin accelerated at 3 m/s^2 towards N-E at an angle of 50° off E direction over a friction less horizontal surface. The acceleration is caused by three horizontal forces of F1= 10 N (S-W 30° off W), F2 = 20 N towards N. What is the third force F3 in unit vector notation and in magnitude angle notation?

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
ACCESS MORE
EDU ACCESS