# How to decide which forces that works on the acceleration?

## The exercise asks about finding the acceleration to the box if it has a $m = 2.35 k g$. But I'm kinda unsure about how one should think about calculating the forces on the box. Since the force is neither vertical or horisontal, but rather diagonal. I am guessing that Newton's second law will be the right approach to this exercise, ∑F=ma. But how do I decide the forces here, since the force is diagonal? If someone could help me with a logical explaination and not only the answer I would really appreciate it.

Jul 21, 2018

$a = 3.69 \frac{m}{s} ^ 2$ rounded to 3 sig figs

#### Explanation:

When looking at a diagonal force, the angle can be used to break down the diagonal force into the horizontal and vertical components of that force, using sine and cosine.

sin30°=x/10

$x = 5$, so there is a vertical force of 5 newtons.

cos30°=x/10

$x = 8.66$ rounded to 3 sig figs, so there is a horizontal force of 8.66 newtons.

Since we want to find the acceleration to the right, we would use the force of $8.66$ newtons in the equation, $F = m a$

$8.66 N = 2.35 k g \cdot a$

$a = 3.69 \frac{m}{s} ^ 2$ rounded to 3 sig figs

Jul 21, 2018

The answer is in the Explanation.

#### Explanation:

The problem does not say anything about friction ... fortunately. It is fortunate because when friction is involved in a more advanced question you will see that the angle complicates matters significantly.

You probably have learned about forming a resultant vector from 2 separate forces working on the same object. What is needed here is to do the reverse -- where the 10 N is the resultant of a vertical force and a horizontal force. Draw the given vector and draw the 2 vectors that could have been the given forces in a "finding the resultant" problem. Then use your trigonometry skills.

$\sin {30}^{\circ} = {F}_{y} / \left(10 N\right)$
Solve for ${F}_{y}$. This force is vertical and is not used in this problem.

$\cos {30}^{\circ} = {F}_{x} / \left(10 N\right)$
Solve for ${F}_{x}$. This force is horizontal and, since the box will move horizontally, is the force to plug into Newton's 2nd.

I hope this helps,
Steve