Question

The figure to the right shows three forces that are acting on a 5.00 kg object. What is the magnitude and direction of the object^?s acceleration? Letter A on the right is 100 it's just a bit cut off.

Letter A on the right is 100 it's just a bit cut off.

Answer 1

To solve this problem you have to know about First Newton's Law

Step 1.

Definition: Every body continues in its state of rest or uniform rectilinear motion unless it is compelled to change its state by forces printed on it. This law states , therefore, that a body can not change alone its initial state, either at rest or in uniform motion unless a force or a number of forces whose resultant is applied is not zero.

   $\vec{F}=m.\vec{a}$

Step 2.

ANSWERS:

Force A:

$\vec{A}= 100N\times (\cos 30\hat{i}+\sin 30\hat{j})\rightarrow \vec{A}=86.60\hat{i}+50\hat{j}$

Force B:

$\vec{B}= 80N\times (\cos 30\hat{i}-\sin 30\hat{j})\rightarrow \vec{B}=69.28\hat{i}-40\hat{j}$

Force C:

$\vec{C}= 40N\times (-\cos 53\hat{i}-\sin 53\hat{j})\rightarrow \vec{C}=-24.07\hat{i}-31.94\hat{j}$

Now applying the first newton's law, we have the following

$\vec{a}=\frac{\sum \vec{F}}{m}$

$\vec{a}=\frac{(86.60\hat{i}+50\hat{j})+(69.28\hat{i}-40\hat{j})+(-24.07\hat{i}-31.94\hat{j})}{5kg}$

$\vec{a}=\frac{(86.60\hat{i}+69.28\hat{i}-24.07\hat{i})+(50\hat{j}-40\hat{j}-31.94\hat{j})}{5}$

$\vec{a}=\frac{131.81\hat{i}-21.94\hat{j}}{5}\rightarrow \vec{a}=26.36\hat{i}-4.38\hat{j}$

Magnitude: $a=\sqrt{(26.36)^{2}+(-4.38)^{2}}\rightarrow a=26.72m/s^{2}$

Direction: $\theta =\tan^{-1}\frac{-4.38}{26.36}\rightarrow \theta =-9.43$ or $\theta =360-9.43\rightarrow \theta =350.5$

If you have any question please let me know in the comments