Calculate Acceleration of 6.90 kg Mass with F1 & F2 Vectors

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Two forces, F1 = -8.20i + 4.30j and F2 = 7.10i + 4.60j, act on a 6.90 kg mass, requiring the calculation of acceleration. The user attempted to add the vectors and apply F=ma but encountered errors in their calculations. It was suggested to draw a diagram to visualize the forces and clarify the relationship between them. The correct approach involves calculating the resultant force and then dividing by the mass to find acceleration, without complicating the process with angles or gravitational force. Simplifying the calculations will yield the correct acceleration value.
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Two forces
F1 = -8.20i + 4.30j and
F2 = 7.10i + 4.60j
are acting on a mass of m = 6.90 kg. The forces are measured in Newtons. What is the magnitude of the object's acceleration?

i tried to add the vectors and using the f=ma to find acceleration but the answer is wrong.

please help
thnx
 

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Draw a picture!

Hi mischaoc! Welcome to PF! :smile:

It always helps to draw a diagram first (just roughly), to see what's going on.

Draw the i and j axes, and three lines with arrows on to represent the two forces and a guess as to the acceleration.

Do you know the relationship between those three lines?
 
Have you neglected to include the force of the mass acting downwards due to gravity? I know that is a mistake that my sister has made when completing similar problems.
 
nice diagram! and colour-coded … I'm impressed!

Hi mischaoc!

From the way you've drawn the diagram, it seems to me that you've worked out what the rule is. :smile:

What's worrying you?

mischaoc said:
i tried to add the vectors and using the f=ma to find acceleration but the answer is wrong.

Show us your working, and the right answer (actually, you should have done that originally).

(and ignore mike's sister - I always ignore mine! :smile:)
 
my work

i added the vectors and gor F3=-1.1i+8.9j
then i calculated the magnitude of
F3=\sqrt{(-1.1)^2+(8.9)^2}=8.9 N

i found the angle \alpha=tan-1 \frac{8.9}{-1.1}
so \alpha=97.5
using the F=ma formula:
8.9*cos(97.5)=6.9*a
a=-0.168
but it's wrong.
 
mischaoc said:
\sqrt{(-1.1)^2+(8.9)^2}=8.9 N

Aha! … now, that's not right, is it? :frown:

You see - that is why the forum rule is that you show what you've done!

Is it ok now? :smile:
 
it's F3=8.98 N
but it is still wrong
 
mischaoc said:
it's F3=8.98 N

Actually, I make it 8.968 (or 8.97).

mischaoc said:
using the F=ma formula:
8.9*cos(97.5)=6.9*a

Why did you put the angle in?

(Are you thinking that j is vertical, and that somehow the weight is involved? That's not what the question says.)

You have the size and direction of the force.

So you divide by m to get the size and direction of the acceleration. That's all!

Don't make it more complicated! :smile:
 
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