Solve Force on Inclined Plane: 30 Degrees & 2.0 m/s Acceleration

[rkslperez04]

My friend and I are working on a worksheet together and she totally lost me and I would like to know how she got her info..

Can you help sifted through her stuff?

Here is the question:

A 2.kg block on a frictionless plane inclined at 30 degrees from horizontal has a force F acting on it upward parallel to the incline. The block is accerlating downward at 2.0 m/s. Find F.

Now here is my work:
I assumed it was the F=ma eqation

F = 2kg*2.0 = 4N

I know its not that easy and cosine of 30 degress comes in somewhere..


Here is her work she sent me in an email:

Now you know the force down the inclined plane. On top of that, you know that there is a force acting up the inclined plane of F. You know that:

(2kg)*g*cos(60 degrees) - F = (2kg)*(2 m/s/s)

That is, assuming that "downward" is the positive direction, then (2kg)*g*cos(60 degrees) is a positive force and F is a negative force. If you sum the forces, you get the net force. That net force is what is causing the 2 m/s/s acceleration. The magnitude of the net force is (2kg)*(2 m/s/s), which is 4 Newtons. Now we have enough to solve for F:

F = (2kg)*g*cos(60 degrees) - (2kg)*(2 m/s/s)
= (2kg)*g*cos(60 degrees) - 4N
= (2kg)*g*(0.5) - 4N
= (1kg)*g - 4N

Now, if you assume that g=9.81m/s/s, then you get:

F = 9.81N - 4N = 5.81N

Thus, the force pulling the block up the hill is 5.81N. Since it is less than the force pulling it down the hill (which is due to gravity), then the block accelerates down the hill at 2.0 m/s/s.

^okokokko.. where did 60 degrees come from and what does the F stand for?

She totally lost and I hate to ask her to clarify again since I missed it the first and she put so much work into emailing me... can you help?

 

[neutrino]

where did 60 degrees come from

Set you x-axis parrallel to the incline and find the componenet of the force due to gravity along the incline. Also, remember, that sin(x) = cos(90-x).

what does the F stand for?

Look at the question you provided.

 

[rsk]

The force F is acting upwards, and the block is accelerating downwards, so it's not the simplest case of F=ma.

Think about the force accelerating the block down - it comes from the weight. So, calculate the compoenent of the weight acting along the slope.

Then you know that the difference between this force, and the force F, gives you the resultant you need to produce the 2m/s^2 acceleration.

 

[radou]

Well, the result doesn't seem correct. There are two forces acting on the box in the direction of the incline - the component of gravity along the incline and the force F. Their directions are opposite, so you have F - G*sin(30) = m*a. You can find the force F easily out of this equation.

 

[Zeno's Paradox]

Your friend is correct .But I understand your consternation.

^okokokko.. where did 60 degrees come from

It maybe helpful if you draw a Free-Body-Diagram and it should be clear where the 60 degrees come from. Note also that 60º is the complementary angle of 30º (because 30º + 60º = 90º), which is the slope of the incline.

and what does the F stand for?

Read again the problem, please:

A 2.kg block on a frictionless plane inclined at 30 degrees from horizontal has a force F acting on it upward parallel to the incline. The block is accerlating downward at 2.0 m/s. Find F.

 

[rsk]

Where does the 60 deg come from?

She's just writing cos60 instead of sin30, that's all. Same thing.

 

[rsk]

Ok, Ok, too many cooks and all that. I´ll back out here.

 

[HallsofIvy]

My friend and I are working on a worksheet together and she totally lost me and I would like to know how she got her info..

Can you help sifted through her stuff?

Here is the question:

A 2.kg block on a frictionless plane inclined at 30 degrees from horizontal has a force F acting on it upward parallel to the incline. The block is accerlating downward at 2.0 m/s. Find F.

Now here is my work:
I assumed it was the F=ma eqation

F = 2kg*2.0 = 4N

I know its not that easy and cosine of 30 degress comes in somewhere..

F, you are told, is upward but the block is moving downward! There must be another force. Do you think it might be gravitational force? The 4 N is the "net" force, the sum of the two opposed forces. What is the component of gravitational force down the incline?

Here is her work she sent me in an email:

Now you know the force down the inclined plane. On top of that, you know that there is a force acting up the inclined plane of F. You know that:

(2kg)*g*cos(60 degrees) - F = (2kg)*(2 m/s/s)

That is, assuming that "downward" is the positive direction, then (2kg)*g*cos(60 degrees) is a positive force and F is a negative force. If you sum the forces, you get the net force. That net force is what is causing the 2 m/s/s acceleration. The magnitude of the net force is (2kg)*(2 m/s/s), which is 4 Newtons. Now we have enough to solve for F:

F = (2kg)*g*cos(60 degrees) - (2kg)*(2 m/s/s)
= (2kg)*g*cos(60 degrees) - 4N
= (2kg)*g*(0.5) - 4N
= (1kg)*g - 4N

Now, if you assume that g=9.81m/s/s, then you get:

F = 9.81N - 4N = 5.81N

Thus, the force pulling the block up the hill is 5.81N. Since it is less than the force pulling it down the hill (which is due to gravity), then the block accelerates down the hill at 2.0 m/s/s.

^okokokko.. where did 60 degrees come from and what does the F stand for?

She totally lost and I hate to ask her to clarify again since I missed it the first and she put so much work into emailing me... can you help?

F stands for exactly what you were told in the problem: "a force F acting on it upward parallel to the incline".

As for where the 60 degrees comes from, I would have written the component of gravitational force along the incline as (2)(9.81) sin(30).
Of course, sin(30)= cos(90).

 

[rkslperez04]

WOW... went to the store and came back to all this! THANKS

Following what you said.. I came up with the following

F (Net Force) = MA

So Force would be the F - mg sin 30.

Rewrite it:

F - mg sin 30 = ma

rearrange

F = m(g sin 30 -a)
F = 2.0 kg(4.9 - 2.0)
F = 2.0 kg * 2.9
F = 5.8 N



GOT IT.. thanks for clearing that up!