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#.

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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.

2 Answers
Jul 21, 2018

Answer:

#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

Jul 21, 2018

Answer:

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