Magnitude of acceleration hwk check

[Ashley1nOnly]

Homework Statement

I wrote down in picture [/B]

Homework Equations

F=ma

The Attempt at a Solution

Worked out in picture [/B]

 

[slider142]

Homework Statement

I wrote down in picture [/B]

Homework Equations

F=ma

The Attempt at a Solution

Worked out in picture [/B]

North, South, East, and West, unless stated otherwise, are directions parallel to the (locally flat) surface of the Earth. In your diagram, you have the second horizontal force, whose direction is described both as being horizontal and being "62° North of East" pointing upwards, away from the ground, with an angle of 62° to the normal to the ground, and partially acting against the vertical force of weight.That direction, normal to the ground, is not called North. It is either called "the normal to the surface", perpendicular to the ground, or zenith.
The given description, however, is meant to convey a direction that is 62° in the northward direction, starting from the direction of cardinal West. There is no vertical component described for that particular force.
As there are no forces described acting vertically, we can assume the normal force of the ground on the block and the weight of the block add up to a net vertical force of 0. So your free body diagram should only have the two horizontal forces described.

 

[Ashley1nOnly]

North, South, East, and West, unless stated otherwise, are directions parallel to the (locally flat) surface of the Earth. In your diagram, you have the second horizontal force, whose direction is described both as being horizontal and being "62° North of East" pointing upwards, away from the ground, with an angle of 62° to the normal to the ground, and partially acting against the vertical force of weight.That direction, normal to the ground, is not called North. It is either called "the normal to the surface", perpendicular to the ground, or zenith.
The given description, however, is meant to convey a direction that is 62° in the northward direction, starting from the direction of cardinal West. There is no vertical component described for that particular force.
As there are no forces described acting vertically, we can assume the normal force of the ground on the block and the weight of the block add up to a net vertical force of 0. So your free body diagram should only have the two horizontal forces described.

So when I take those off. My x components are still right. When I sum of all my components in the x direction I should have f1-f2cos(29)=m•a

 

[slider142]

So when I take those off. My x components are still right. When I sum of all my components in the x direction I should have f1-f2cos(29)=m•a

Not quite. Your diagram has the second vector's direction as 62° West of the North cardinal direction. The description in the problem, however, says the vector should be 62° North of West, which, in your diagram, means we start on the West direction (the leftwards pointing horizontal axis in your diagram) and we move 62° clockwise towards the North direction. In other words, the net East-West force should be F₁ - F₂cos(62°).

 

[Ashley1nOnly]

Not quite. Your diagram has the second vector's direction as 62° West of the North cardinal direction. The description in the problem, however, says the vector should be 62° North of West, which, in your diagram, means we start on the West direction (the leftwards pointing horizontal axis in your diagram) and we move 62° clockwise towards the North direction. In other words, the net East-West force should be F₁ - F₂cos(62°).

Ohhhhhh. I see that. If it said 62 west of north. Then I would draw my angle 62 degrees from my north direction. Thanks.

Now we have (f1-f2cos(62))/m =a. We have solved our problem.

 

[slider142]

Ohhhhhh. I see that. If it said 62 west of north. Then I would draw my angle 62 degrees from my north direction. Thanks.

Now we have (f1-f2cos(62))/m =a. We have solved our problem.

That's it! However, this is only the East-West component of the acceleration vector. There is also a component of net force in the North direction, which gives us a component of acceleration in that direction as well. Once you have both components, you can find the magnitude of the acceleration vector in the usual way.

 

[Ashley1nOnly]

That's it! However, this is only the East-West component of the acceleration vector. There is also a component of net force in the North direction, which gives us a component of acceleration in that direction as well. Once you have both components, you can find the magnitude of the acceleration vector in the usual way.

The acceleration in the y direction would be zero so that shouldn't count right.

F2sin(62)=ma
But a =o

 

[slider142]

The acceleration in the y direction would be zero so that shouldn't count right.

F2sin(62)=ma
But a =o

Why do you believe that the acceleration in the y direction should be zero ?

 

[Ashley1nOnly]

Why do you believe that the acceleration in the y direction should be zero ?

Because the box is not accelerating in the y direction. Is only moving along the x direction as it moves across the frictionless table top.
So since it's not moving in the y direction the acceleration is 0

 

[slider142]

Because the box is not accelerating in the y direction. Is only moving along the x direction as it moves across the frictionless table top.

Hmm, this information is not given in the problem. What information did you use to conclude that the box is only moving in the x direction ? Remember the x and y axes are both horizontal if they are aligned with the cardinal directions. The vertical axis, along which the weight of the box is aligned, would be the z axis in this case.

 

[Mister T]

In your original diagram you chose the y-axis to be vertical, yet it's stated that the two given forces are in horizontal directions.

Best to re-draw your diagram and realize that the forces ##m \vec{g}## and ##\vec{F}_N## don't belong there because ##m \vec{g}## points downward and ##\vec{F}_N## points upward. They lie along the z-axis, given that the xy-plane is horizontal.

 

[Ashley1nOnly]

Hmm, this information is not given in the problem. What information did you use to conclude that the box is only moving in the x direction ? Remember the x and y axes are both horizontal if they are aligned with the cardinal directions. The vertical axis, along which the weight of the box is aligned, would be the z axis in this case.

This is just a 2d problem so the z axis wouldn't be included. And based on the other examples we have done the a=0 with of course the block being on an object. If the block was being pulled in by a pulley then that would be a different case when a DIDNT equal zero

 

[Ashley1nOnly]

This is just a 2d problem so the z axis wouldn't be included. And based on the other examples we have done the a=0 with of course the block being on an object. If the block was being pulled in by a pulley then that would be a different case when a DIDNT equal zero

We never do anything in 3d in my class either.

 

[Ashley1nOnly]

In your original diagram you chose the y-axis to be vertical, yet it's stated that the two given forces are in horizontal directions.

Best to re-draw your diagram and realize that the forces ##m \vec{g}## and ##\vec{F}_N## don't belong there because ##m \vec{g}## points downward and ##\vec{F}_N## points upward. They lie along the z-axis, given that the xy-plane is horizontal.

Yeah the only forces that should be there are:

F1 point in the east direction as dram
And
F2 point in a 62 degree angle north of west which has to be broken into its x an y components which I have done.

But the y wouldn't matter because the box is not accelerating in the y direction which would make it zero.

 

[Mister T]

This is just a 2d problem so the z axis wouldn't be included.

I agree. Which is why I said this ...

Best to re-draw your diagram and realize that the forces ##m \vec{g}## and ##\vec{F}_N## don't belong there [...]

 

[Mister T]

But the y wouldn't matter because the box is not accelerating in the y direction which would make it zero.

There is no acceleration in the vertical direction, but the y-axis is not vertical!

 

[Ashley1nOnly]

I agree. Which is why I said this ...

Okay I solved for a for my final answer

 

[Mister T]

Why do you show the ice rink below the puck? You are looking down on the rink from above, so all you'll see is the round shape of the puck.

I ask this, not because I want to see a better drawing, but because you're still not getting the right answer and I'm trying to get you to see why so you can fix it.

By the way, the northward direction is not up. It's a direction that lies in the same horizontal plane as east and west.

I realize that nothing you do in your class is in three dimensions, but you still need to be able to orient the two-dimensional plane you are working in. That plane could be vertical or it could be horizontal. In this case it's horizontal but you are thinking of it, and drawing it, as if it's vertical.

 

[Ashley1nOnly]

Why do you show the ice rink below the puck? You are looking down on the rink from above, so all you'll see is the round shape of the puck.

I ask this, not because I want to see a better drawing, but because you're still not getting the right answer and I'm trying to get you to see why so you can fix it.

By the way, the northward direction is not up. It's a direction that lies in the same horizontal plane as east and west.

I realize that nothing you do in your class is in three dimensions, but you still need to be able to orient the two-dimensional plane you are working in. That plane could be vertical or it could be horizontal.

I thought we were looking at the block from the side, like if it were sitting on a table and we were trying to calculate the forces then. Then the way I drew the vectors would be right.Is the answer wrong like the numerical value.

 

[Ashley1nOnly]

I thought we were looking at the block from the side, like if it were sitting on a table and we were trying to calculate the forces then. Then the way I drew the vectors would be right.Is the answer wrong like the numerical value.

I would say the plane was horizontal which is why I drew the little graph showing that the y was up and the was horizontal

 

[Mister T]

If the plane is horizontal, up and down are perpendicular to the plane and will not appear as directions in that plane.

You do not have the right answer, but you are now on the right track. And the way you show and do your work is excellent, too.

 

[Ashley1nOnly]

I would say the plane was horizontal which is why I drew the little graph showing that the y was up and the was horizontal

No it's going to be a vertical xy plane to make what I suggested true

 

[Ashley1nOnly]

I thought we were looking at the block from the side, like if it were sitting on a table and we were trying to calculate the forces then. Then the way I drew the vectors would be right.Is the answer wrong like the numerical value.

Why do you show the ice rink below the puck? You are looking down on the rink from above, so all you'll see is the round shape of the puck.

I'm looking at the puck from the side not on top for this problem. But I do understand what you were trying to say now that I go back and read your statements

 

[Mister T]

If you're looking at a vertical plane the two horizontal forces acting on the puck cannot both appear in that plane. For example, if one of them is towards the east (to your right) the other cannot be drawn in that same plane, it would be at an angle to that plane.

Like I said, best to use a horizontal plane.

 

[Ashley1nOnly]

If you're looking at a vertical plane the two horizontal forces acting on the puck cannot both appear in that plane. For example, if one of them is towards the east (to your right) the other cannot be drawn in that same plane, it would be at an angle to that plane.

Like I said, best to use a horizontal plane.

Would that change my answer for my acceleration??

 

[Mister T]

Yes, that's the idea.

Go back and read the problem again. Think of the two forces given in the statement of the problem as forces applied by strings tied to the puck. Those two forces combine to form a net force. That net force, divided by the mass, gives you the acceleration.

 

[Ashley1nOnly]

Yes, that's the idea.

Go back and read the problem again. Think of the two forces given in the statement of the problem as forces applied by strings tied to the puck. Those two forces combine to form a net force. That net force, divided by the mass, gives you the acceleration.

X
F1-f2cos(62)=ma
How would you draw it or write it if you can give a picture

 

[Ashley1nOnly]

Yes, that's the idea.

Go back and read the problem again. Think of the two forces given in the statement of the problem as forces applied by strings tied to the puck. Those two forces combine to form a net force. That net force, divided by the mass, gives you the acceleration.

Ok I went back and looked at it from you point of view and totally get what you were talking about with the whole puck thing. The normal force and gravitational force are in the z direction. Okay I got that point now. The xy plane is horizontal to the z plane which makes sense. So there should be movement/ an acceleration in the y direction. From that view

X
F1-f2cos(62)=ma

Y
F2sin(62)=ma
Rewriting
X
(F1-f2cos(62))/m =a

Y

(F2sin(62))/m =a

Now how do we combine them

 

[Ashley1nOnly]

Ok I went back and looked at it from you point of view and totally get what you were talking about with the whole puck thing. The normal force and gravitational force are in the z direction. Okay I got that point now. The xy plane is horizontal to the z plane which makes sense. So there should be movement/ an acceleration in the y direction. From that view

X
F1-f2cos(62)=ma

Y
F2sin(62)=ma
Rewriting
X
(F1-f2cos(62))/m =a

Y

(F2sin(62))/m =a

Now how do we combine them

The puck is not moving in the z direction which would make it zero

 

[Mister T]

Just like you drew it in your original picture, except you would omit ##m \vec{g}## and ##\vec{F}_N##.

 

[Ashley1nOnly]

Just like you drew it in your original picture, except you would omit ##m \vec{g}## and ##\vec{F}_N##.

Yep and I see why now, thanks so much for not giving me the answers and making me work for it.

Am I right now with my mind of thinking

 

[Mister T]

X
(F1-f2cos(62))/m =a

Y

(F2sin(62))/m =a

Now how do we combine them

The first equation is actually equal to a_x, the second is equal to a_y.

Do you know how to combine them to find the magnitude of ##\vec{a}##?

Alternatively, you could have combined F_x and F_y to find the magnitude of ##\vec{F}## first. And then divided that by m to find the magnitude of ##\vec{a}##.

 

[Ashley1nOnly]

Just like you drew it in your original picture, except you would omit ##m \vec{g}## and ##\vec{F}_N##.

Then going on with the problem to finish it

I have everything summed up
F(net)=(9-8cos(62))i +(F2sin(62))j
=5.24i + 7.06j
Take the magnitude of it
Sqrt( (5.24)^2+(7.06)^2)
=8.8
Which gave me the next force. Now in order to find the acceleration I know that
F(net)=ma

A= f(net)/m
A=8.8/3.0
A=2.93

 

[Mister T]

Nice!

 

[Ashley1nOnly]

Nice!

Thanks so much.
WOOT woot