Calculating Acceleration with Multiple Forces Acting on an Object

[SDTK]

Homework Statement

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. Homework Equations

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)?

 

[ObjectivelyRational]

I get 1.0577 m/s^2

 

[SDTK]

I get 1.0577 m/s^2

Thank you! :-)

 

[haruspex]

I get 1.0577 m/s^2

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.

 

[SDTK]

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.

thank you

 

[SDTK]

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

 

[SDTK]

thank you

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?

 

[TomHart]

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

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.

 

[gneill]

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.

 

[SDTK]

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

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.

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

 

[SDTK]

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.

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

 

[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.

 

[gneill]

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).

 

[haruspex]

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).

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

 

[haruspex]

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

View attachment 108525

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

 

[gneill]

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

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.

 

[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?

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.

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.

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.

 

[SDTK]

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

63 degrees

 

[gneill]

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.

 

[haruspex]

63 degrees

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.

 

[SDTK]

thank you, thank you, thank you! I recalculated the F(net) x, getting 52 N.
The illustration you shared helped! :-)

 

[SDTK]

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.

thank you! I did not know the cosine rule. I looked it up, and see how works, and it gives me the right answer! :-)

 

[SDTK]

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

View attachment 108530

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.

gneil,
thank you! I posted a reply, but it looks like it went as a separate posting instead of a direct reply. I see my error in the x_ component calculation.
I (now) have 52N for F(net)x.
Thank you also for the notes about the Cartesian components for the net force vector!

 

[gneill]

gneil,
thank you! I posted a reply, but it looks like it went as a separate posting instead of a direct reply. I see my error in the x_ component calculation.
I (now) have 52N for F(net)x.
Thank you also for the notes about the Cartesian components for the net force vector!

You're welcome.

Looks like you now have two methods in your toolkit that you can use to add vectors.

 

[SDTK]

@gneill, @haruspex, @ObjectivelyRational, @TomHart; Thank you all for the help!
I understand how to do this now.