Question

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 = 2.35kg`.

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.

Answer 1

`a=3.69 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=ma`

`8.66N=2.35kg*a`

`a=3.69 m/s^2` rounded to 3 sig figs

Answer 2

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.

`sin30^@ = F_y/(10 N)`
Solve for `F_y`. This force is vertical and is not used in this problem.

`cos30^@ = F_x/(10 N)`
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