# Magnitude of acceleration problem

1. Dec 2, 2007

### zjm7290

[SOLVED] Magnitude of acceleration problem

1. The problem statement, all variables and given/known data
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 7.7 N, acting 54° north of west. What is the magnitude of the body's acceleration?

2. Relevant equations
I know that in order to get net force, you would need to add all force vectors in the x and y direction giving you total forces in x direction then total forces in y direction. Then I would use the equation given by Newton's second law F (net)=m*a (net)

3. The attempt at a solution
I set up a drawing showing the 9.0N force in the east direction on the x axis, then the 7.7N force in the west direction 54 degrees above the x axis. I added all vectors in the x direction getting 9.0N + 7.7N*sin(54) which equals 15.229N in x direction. Then I added all vectors in y direction getting 0N + 7.7N*cos(54) which equals 4.256N in the y direction. After this, i used F=m*a to get acceleration in both x and y directions. This assignment uses web-assign (on-line pragram) which gives you 5 attempts at a solution before it counts it wrong. This assignment counts for a test grade and I'm on my final attempt. Any guidance or help would be greatly appreciated! Thanks in advance.....

2. Dec 2, 2007

### G01

It seems to me that the only two problems you have here are:

1)Trig functions. You are using the wrong trig function in each case. Set up the triangles formed by the 7.7N vector and its components and you should see that you need to use cos(54) to find the x component and the sin(54) to find the y component.

2) Remember that if you let the 9.0N force be positive, the x component of the 7.7N force will have to be negative, since it points the opposite way.

3. Dec 2, 2007

### zjm7290

so in the x direction I would have 9.0N + -7.7N*cos(54) and in the y direction I would have 0N + 7.7N*sin(54)? Or would I apply the negative sign after I figure out 7.7N*cos(54) since thats the total force in that direction with reference to the x axis

4. Dec 2, 2007

### G01

What you have written there is correct.

5. Dec 2, 2007

### zjm7290

so I now get F (net) in x direction = 4.474N and F(net) in y direction = 6.229N, I then use F=ma to get acceleration for both x and y directions so therefore the magnitude of acceleration in the system is a(net) in x direction + a(net) in y direction

6. Dec 2, 2007

### G01

remember that $$\vec{a}_{net x}$$ and $$\vec{a}_{net y}$$ are vectors and must be added like vectors. Otherwise, that is correct so far.

7. Dec 2, 2007

### zjm7290

so you mean that since the 7.7N force is in the neg x direction, the acceleration in that same direction would also be negative?

8. Dec 2, 2007

### G01

Yes, that is correct, but what I was trying to point out is that a(netx) point perpendicular to a(nety), so you can't just add them like real numbers. You have to use the Pythagorean Theorem.

9. Dec 2, 2007

### zjm7290

so basically the net accelelration is the vector quantity of where the end of one force connects to the other or the last line that would complete the triangle that would show the total force vectors in the system?

10. Dec 2, 2007

### G01

Yes, it is the vector sum of the x and y accelerations.

11. Dec 2, 2007

### zjm7290

thanks so much for all the help!

12. Dec 2, 2007

No problem!