# Two forces act on an object

1. Nov 4, 2016

### SDTK

1. The problem statement, all variables and given/known data
Two forces act on a 55 kg object. One has a magnitude of 65 N directed 59 degrees clockwise from the pos. x axis. The other has magnitude 35 N at 32 o clockwise from the pos. y axis.What is the acceleration of the object?

The answer is given as 1.1 m/ s squared,
(I got a similar answer 1.0 m/s squared.) , but not confident that I approached the solution correctly.

Thanks in advance for any help :-)

2. Relevant equations

3. The attempt at a solution
-- Is the correct approach to find the sum of: all x components, and y components of the forces, (65N, 35N, and weight)?

Last edited by a moderator: Nov 4, 2016
2. Nov 4, 2016

### ObjectivelyRational

I get 1.0577 m/s^2

3. Nov 4, 2016

### SDTK

Thank you! :-)

4. Nov 4, 2016

### haruspex

I confirm that.
@SDTK, if that is not what you got, please post your detailed working. Most likely you rounded some intermediate answer, keeping insufficient precision.

5. Nov 4, 2016

thank you

6. Nov 4, 2016

### SDTK

I also looked at doing vector addition, but it doesn't give a right triangle.

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7. Nov 5, 2016

### SDTK

I also looked at vector addition, but I would not get a right triangle. I'm thinking that vector addition would not be appropriate--- Is that correct?

8. Nov 5, 2016

### TomHart

SDTK, you wrote:
ΣFx = sin(32)(35) + cos(59)(65) = max
620.19 = max

How did you get from the first line to the second line?

Edit: It looks like you may have multiplied those 2 terms, rather than add them.

Last edited: Nov 5, 2016
9. Nov 5, 2016

### Staff: Mentor

The problem states that two forces act on the object. Gravity was not one of them.

Cartesian vector components add in quadrature: square root of the sum of the squares.

10. Nov 5, 2016

### SDTK

Not sure what I did, thank you! for pointing it out :-)

11. Nov 5, 2016

### SDTK

gneill
Thank you!
Is this what you are telling me? (work shown below)
(I'm concerned because I get 0.99 m/s^2, which is different from the answers of @haruspex 1.0577m/s^2, @ObjectivelyRational 1.0577m/s^2, and the answer given with the problem 1.1m/s^2.)
Thanks

12. Nov 5, 2016

### ObjectivelyRational

Find x component of the net force, then find it in Y direction. THESE are of course 90 degrees wrt each other. Calculate the magnitude of the total net force. then calculate a from F and m.

13. Nov 5, 2016

### Staff: Mentor

No. The given forces won't in general form a right angle triangle (and they don't in this case). You need to sum their components.

What I was getting at was, once you've summed the individual x and y components of the vectors to form the resultant vector's components, then the magnitude of that resultant is found by summing its x and y components. You still need to find the components of the resultant vector using the method that you used in post #6 (only forget the gravitational force since it plays no role here).

14. Nov 5, 2016

### haruspex

Isn't it more direct just to draw the vector triangle and apply the cosine rule?

15. Nov 5, 2016

### haruspex

In the triangle of forces you drew, what is the angle between the 65N and the 35N? Just a bit of elementary geometry needed.

16. Nov 5, 2016

### Staff: Mentor

Yes, if you have the angle between the forces. In general though I expect a student to be faced with summing multiple forces to find a resultant. Doing them one at a time via the cosine rule could be tedious.

17. Nov 5, 2016

### SDTK

Thank you!
I found
f net x to be 67.92N sin(32)35N + cos(59)65N
f net y to be -26.03 N cos(32)35N - sin(59)65N

If I draw a vector, magnitude of 68N on the positive x axis, and a vector magnitude 26N on the negative y axis,... then draw a vector between the two that meets the segment representing the vector sum, is that a vector representing the total net force?

18. Nov 5, 2016

### SDTK

63 degrees

19. Nov 5, 2016

### Staff: Mentor

Cartesian components of a vector form a rectangle with the sides paralleling the coordinate axes and the vector itself forming the diagonal:

Note that the components $f_x$ and $f_y$ of vector $f$ are at right angles to each other and are parallel to the axes.

The magnitude of f is $|f| = \sqrt{f_x^2 + f_y^2}$

I don't agree with the value of the x_component that you calculated in your post #17. Can you give more details of that calculation? The y_component looks okay.

20. Nov 5, 2016

### haruspex

Ok, so you have a choice of approaches. For a simple resultant of two vectors, you can use the angle between them (when drawn in the nose-to-tail format of the force triangle) and apply the cosine rule, or for multiple applied forces you can reduce each to its X and Y components and find the sum along each axis.