Question

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?

Answer 1

Explanation:

The net force is calculated as follows.

        $F_{net} = F_{1} + F_{2} + F_{3}$

Also,     $F_{net}$ = ma

                         = $2 \times 3$

                         = 6 N

       = $6 Cos 50 \hat{i} + 6 Sin 50 \hat{j}$

       = $3.86 \hat{i} + 4.6 \hat{j}$

  $F_{1} = -10 Cos (30)\hat{i} - 10 Sin (30)\hat{j}$

               = $-8.66\hat{i} - 5\hat{j}$

  $F_{2} = 20\hat{j}$

$3.86 \hat{i} + 4.6 \hat{j}$ = $(-8.66\hat{i} - 5\hat{j}) + 20\hat{j} + F_{3}$

         $F_{3} = 12.52\hat{i} - 10.4\hat{j}$

                     = 16.3 N

Thus, we can conclude that the third force $F_{3}$ in unit vector notation is 16.3 N and magnitude angle notation is $39.7^{o}$.