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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.Ashley1nOnly said:Homework Statement
I wrote down in picture [/B]Homework Equations
F=ma
The Attempt at a Solution
Worked out in picture [/B]
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•aslider142 said: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.
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 said: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
Ohhhhhh. I see that. If it said 62 west of north. Then I would draw my angle 62 degrees from my north direction. Thanks.slider142 said: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°).
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 said: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.
The acceleration in the y direction would be zero so that shouldn't count right.slider142 said: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.
Why do you believe that the acceleration in the y direction should be zero ?Ashley1nOnly said:The acceleration in the y direction would be zero so that shouldn't count right.
F2sin(62)=ma
But a =o
slider142 said:Why do you believe that the acceleration in the y direction should be zero ?
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.Ashley1nOnly said: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.
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 zeroslider142 said: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.
We never do anything in 3d in my class either.Ashley1nOnly said: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
Yeah the only forces that should be there are:Mister T said: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 said:This is just a 2d problem so the z axis wouldn't be included.
Mister T said:Best to re-draw your diagram and realize that the forces ##m \vec{g}## and ##\vec{F}_N## don't belong there [...]
Ashley1nOnly said:But the y wouldn't matter because the box is not accelerating in the y direction which would make it zero.
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.Mister T said: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 would say the plane was horizontal which is why I drew the little graph showing that the y was up and the was horizontalAshley1nOnly said: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.
No it's going to be a vertical xy plane to make what I suggested trueAshley1nOnly said: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
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.Ashley1nOnly said: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.
Would that change my answer for my acceleration??Mister T said: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.
XMister T said: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 viewMister T said: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.
The puck is not moving in the z direction which would make it zeroAshley1nOnly said: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
Yep and I see why now, thanks so much for not giving me the answers and making me work for it.Mister T said:Just like you drew it in your original picture, except you would omit ##m \vec{g}## and ##\vec{F}_N##.
Ashley1nOnly said:X
(F1-f2cos(62))/m =a
Y
(F2sin(62))/m =a
Now how do we combine them
Then going on with the problem to finish itMister T said:Just like you drew it in your original picture, except you would omit ##m \vec{g}## and ##\vec{F}_N##.
Thanks so much.Mister T said:Nice!
The magnitude of acceleration is a measure of how much an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude (size) and direction. It is usually measured in meters per second squared (m/s²).
The magnitude of acceleration can be calculated by dividing the change in velocity (Δv) by the change in time (Δt). This can be represented by the equation a = Δv/Δt. Alternatively, it can also be calculated by multiplying the mass of an object (m) by its acceleration due to gravity (g), which is approximately 9.8 m/s² on Earth.
The magnitude of acceleration is affected by several factors, including the mass of the object, the force applied to the object, and the surface on which the object is moving. For example, a heavier object will require a greater force to accelerate it compared to a lighter object. Additionally, a smoother surface will result in less friction and therefore a greater acceleration.
The magnitude of acceleration is closely related to Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. In other words, the greater the force applied to an object, the greater its acceleration will be, and the more massive an object is, the less it will accelerate under the same force.
Understanding the magnitude of acceleration is crucial in many fields of science and engineering. It allows us to calculate the motion and behavior of objects, such as the trajectory of a projectile or the speed of a moving car. It also helps us design and improve technologies, such as vehicles, roller coasters, and aircraft, by ensuring they can safely accelerate and decelerate within certain limits.