Question

A 2 kg object is subjected to three forces that give it an acceleration −→a = =(8.00m/s2 )ˆi + (6.00m/s2 )ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = =(12.0N)ˆi + (8.00N)ˆj, find the third force.

Answer 1

Answer:

The third force  = $-(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})$

Explanation:

Here we note that, the formula for the resultant force is as follows;

$\Sigma F = \Sigma F_x + \Sigma F_y$

Also

∑F = m×a

Where:

$F_x$ = Force components in the x direction

$F_y$ = Force components in the y direction

∑F = Resultant force vector

m = Mass of the object

a = Acceleration vector ob the object =

$a = 8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j}$

$F_1 = 30.0 \, N \, \hat{i} + 16.0 \, N \, \hat{j}$

$F_2 = 12.0 \, N \, \hat{i} + 8.00 \, N \, \hat{j}$

Therefore, since ∑F = m×a, we have;

$\Sigma F = 2 kg \times (8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j})$

$\Sigma F = (16.00 \, m/s^2 \, \hat{i} + 12.00 \, m/s^2 \, \hat{j})$

Hence from $\Sigma F = \Sigma F_x + \Sigma F_y$, we have;

$F_{x1} + F_{x2} + F_{x3} = 16$

That is 30 + 12 + $F_{x3}$ = 16

∴  $F_{x3}$ = 16 - (30 + 12) = -26

Similarly,

$F_{y1} + F_{y2} + F_{y3} = 12$

Therefore,  $F_{y3}$ = 12 - (16 + 8) = -12

Hence, $F_3 = -26.0 \, N \, \hat{i} - 12.00 \, N \, \hat{j} = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})$

The third force, $F_3, = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})$