Force Diagram (A2 Edexcel)
are the components of the forces that i labelled wrong? if so, then which ones.
i get 549 m but the answer is supposed to be 816 m(3 s.f)
here's the question without any markings:
i get 549 m but the answer is supposed to be 816 m(3 s.f)
here's the question without any markings:
(edited 11 months ago)
Original post by Aleksander Krol
are the components of the forces that i labelled wrong? if so, then which ones.
i get 549 m but the answer is supposed to be 816 m(3 s.f)
here's the question without any markings:
i get 549 m but the answer is supposed to be 816 m(3 s.f)
here's the question without any markings:
It is the angle that you make the mistake.
The lift force is not making at an angle of 20 deg with the horizontal.
Hi, I would recommend you use the snipping tool in Windows 10/11 to take pictures of your question so that we can see clear pictures.
Below are some videos to show you how to use the snipping tool.
https://youtu.be/2O932t8-k2A
https://youtu.be/fWNA2js9JGc
https://youtu.be/FPXTvrY7yc4
Below are some videos to show you how to use the snipping tool.
https://youtu.be/2O932t8-k2A
https://youtu.be/fWNA2js9JGc
https://youtu.be/FPXTvrY7yc4
Original post by Eimmanuel
Hi, I would recommend you use the snipping tool in Windows 10/11 to take pictures of your question so that we can see clear pictures.
Below are some videos to show you how to use the snipping tool.
https://youtu.be/2O932t8-k2A
https://youtu.be/fWNA2js9JGc
https://youtu.be/FPXTvrY7yc4
Below are some videos to show you how to use the snipping tool.
https://youtu.be/2O932t8-k2A
https://youtu.be/fWNA2js9JGc
https://youtu.be/FPXTvrY7yc4
hi, thanks for recommending me that. sorry about the clarity of the image, i took it in a hurry
Original post by Eimmanuel
It is the angle that you make the mistake.
The lift force is not making at an angle of 20 deg with the horizontal.
The lift force is not making at an angle of 20 deg with the horizontal.
would it be possible for you to giude me to do it in the right way.
i'm considering the plane as an object on a ramp like in M1.
are we going to consider a component of lift force in a straight horizontal line pointing to the centre in the leftward direction as a cnetripetal force?
(edited 11 months ago)
Original post by Aleksander Krol
would it be possible for you to giude me to do it in the right way.
i'm considering the plane as an object on a ramp like in M1.
are we going to consider a component of lift force in a straight horizontal line pointing to the centre in the leftward direction as a cnetripetal force?
i'm considering the plane as an object on a ramp like in M1.
are we going to consider a component of lift force in a straight horizontal line pointing to the centre in the leftward direction as a cnetripetal force?
Can you see an alternate angle in blue?
Original post by Eimmanuel
Can you see an alternate angle in blue?
Can you see an alternate angle in blue?
yes
so is it going to be like
image below*
(edited 11 months ago)
Original post by Aleksander Krol
yes
so is it going to be like
image below*
so is it going to be like
image below*
Your lift force is along the wing.
You "should" resolve the lift force using the angle drawn in green.
Original post by Aleksander Krol
are the components of the forces that i labelled wrong? if so, then which ones.
i get 549 m but the answer is supposed to be 816 m(3 s.f)
here's the question without any markings:
i get 549 m but the answer is supposed to be 816 m(3 s.f)
here's the question without any markings:
The equation " L = mg cos20 " is not correct. As Eimmanuel pointed out earlier the angle is not 20.
Also you need to consider the component of centripetal acceleration in the direction of lift.
Original post by thomas.rhett
The equation " L = mg cos20 " is not correct. As Eimmanuel pointed out earlier the angle is not 20.
Also you need to consider the component of centripetal acceleration in the direction of lift.
Also you need to consider the component of centripetal acceleration in the direction of lift.
but why? isn't the case same as this one:
where R = mg x cos(theta)
i just plugged the Lift force in the place of R.
(edited 11 months ago)
Original post by Aleksander Krol
but why? isn't the case same as this one:
where R = mg x cos(theta)
i just plugged the Lift force in the place of R.
where R = mg x cos(theta)
i just plugged the Lift force in the place of R.
No. Because look at the direction of acceleration in the two cases. The acceleration of the particle is perpendicular to R (the normal force) but centripetal acceleration of the plane is NOT perpendicular to the lift.
Original post by Eimmanuel
Your lift force is along the wing.
You "should" resolve the lift force using the angle drawn in green.
You "should" resolve the lift force using the angle drawn in green.
i did the calculation shown in the image, and i got 816 m (3 s.f)
now my only confusion is why the ones i marked in red is not considered?
i'm trying to understand, but i'm not getting it yet.
Original post by Aleksander Krol
i did the calculation shown in the image, and i got 816 m (3 s.f)
now my only confusion is why the ones i marked in red is not considered?
i'm trying to understand, but i'm not getting it yet.
i did the calculation shown in the image, and i got 816 m (3 s.f)
now my only confusion is why the ones i marked in red is not considered?
i'm trying to understand, but i'm not getting it yet.
mg is vertical so has no component in the horizonal direction. It has a component in the opposite direction of lift which you have correctly marked mg cos20. mgsin20 is parallel to the wing.
Original post by thomas.rhett
mg is vertical so has no component in the horizonal direction. It has a component in the opposite direction of lift which you have correctly marked mg cos20. mgsin20 is parallel to the wing.
understood this part now. thanks a bunch for that!!
so now it's like this:
now the main question is about why "Lift=/ mgcos20"
they are opposite to each other. so why should we not do:
Lift= mgcos20°
and get the value for lift instead of finding the value of lift force by doing:
lift x sin(70°) = mg
if i'm not wrong, i believe you've already explained that, so if you don't want to anymore, it's understandable. thank you still for helping!!
Original post by Aleksander Krol
understood this part now. thanks a bunch for that!!
so now it's like this:
now the main question is about why "Lift=/ mgcos20"
they are opposite to each other. so why should we not do:
Lift= mgcos20°
and get the value for lift instead of finding the value of lift force by doing:
lift x sin(70°) = mg
if i'm not wrong, i believe you've already explained that, so if you don't want to anymore, it's understandable. thank you still for helping!!
so now it's like this:
now the main question is about why "Lift=/ mgcos20"
they are opposite to each other. so why should we not do:
Lift= mgcos20°
and get the value for lift instead of finding the value of lift force by doing:
lift x sin(70°) = mg
if i'm not wrong, i believe you've already explained that, so if you don't want to anymore, it's understandable. thank you still for helping!!
When we write these equations, what is it that we are doing? We are applying F = ma. Now if a = 0 in a particular direction, then forces are in equilibrium and resultant force (sum of components of all forces in that direction) must be zero. If a is not zero then forces are not in equilibrium and there is a resultant force equal to ma.
In the direction of lift the forces are not in equilibrium because there is component of centripetal acceleration in the direction of lift given by ((v^2)/r)cos70. So L - mgcos20 is not zero. The correct equation is L - mgcos20 = m((v^2)/r)cos70.
Forces are also not in equilibrium parallel to the wing since there is component of centripetal acceleration along the wing. But lift is perpendicular to the wing so it has no component in this direction, therefore mgsin20 = m((v^2)/r) sin70.
The plane is not accelerating vertically so force are in equilibrium vertically. So Lsin70 - mg = 0.
Forces are not in equilibrium horizontally because of circular motion so Lcos70 = m((v^2)/r).
When solving problems, you can only get 2 equations in this way regardless of how many unknowns you have. You can resolve forces in any direction and the direction perpendicular to that direction. Even if you were to write another equation by resolving forces in a different direction it won't be independent from the first 2.
i think i understood what you were trying to explain all this time. basically we can consider the coordinate axis however we want. but for simplification of calculation, we consider the x axis to be in parallel with the direction of acceleration.
for e.g: for an object in an inclined plane, we consider the x axis to be parallel to the slope becasue it is accerating parallel to the slope:
and for the case of plane banking, the plane has an acceleration perpendicular to the weight force, therefore we consider the x axis to be perpendicular to the weight of the plane and y axis to be parallel to the weight of the plane for easier calculation:
while considering which ever coordinate axis, we have to keep in mind, just because the forces are opposite to each other in the y axis or x axis does not mean that they are equal to each other.
for e.g in the case of an object falling from the inclined plane, obviously the forces in the x axis are not balanced, that's why it is falling or there is a resultant force which is causing this motion (unbanced forces). but the forces are balanced vertically as there is no motion vertically for the object falling from the ramp or inclined plane.
Also, noting that in the case for the plane banking, because we chose the y axis to be parallel to the weight of the plane, we cannot breakdown the weight into component or resolve weight horzontally or vertically, because it is already vertical or parallel to the y axis.
that's why we are only going to breakdown the lift force into components by resolving it vertically and horizontally:
for e.g: for an object in an inclined plane, we consider the x axis to be parallel to the slope becasue it is accerating parallel to the slope:
and for the case of plane banking, the plane has an acceleration perpendicular to the weight force, therefore we consider the x axis to be perpendicular to the weight of the plane and y axis to be parallel to the weight of the plane for easier calculation:
while considering which ever coordinate axis, we have to keep in mind, just because the forces are opposite to each other in the y axis or x axis does not mean that they are equal to each other.
for e.g in the case of an object falling from the inclined plane, obviously the forces in the x axis are not balanced, that's why it is falling or there is a resultant force which is causing this motion (unbanced forces). but the forces are balanced vertically as there is no motion vertically for the object falling from the ramp or inclined plane.
Also, noting that in the case for the plane banking, because we chose the y axis to be parallel to the weight of the plane, we cannot breakdown the weight into component or resolve weight horzontally or vertically, because it is already vertical or parallel to the y axis.
that's why we are only going to breakdown the lift force into components by resolving it vertically and horizontally:
(edited 11 months ago)
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