Tension Free Response QQT Solution (AP Physics 1)
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- Опубліковано 21 жов 2024
- Tension Free Response QQT Solution (AP Physics 1)
This question look at solving for the tension on a string while a ball is at rest, while a ball is accelerating but on a horizontal surface and an inclined plane as well as in a vertical loop. This covers the topic of dynamics and circular motion.
If there is a topic you want me to do leave them in the comments below.
#physicstutor #APPhysics1 #APFreeResponse
Woah, part (d) was intense! I think I watched it 3 times.
Def a good question!
For part d, I found splitting the acceleration into horizontal and vertical components was easier for me than trying to figure out the net force. Then, you can just solve that m*a sub h = T sub h and m*a sub v = mg-T sub v
Do you bud
For part c, why is it mg - Tv, shouldn't it be Tv - mg because the mg force is acting in the negative direction?
Toward the center is always positive
@@themathandphysicstutor Oh ok thank you so much
For Part D, how does the string remain suspended in the air. Wouldn't gravity cause the string to lie on the roof of the car?
because its moving in a circular path so it wants to move tangent to the circle so gravity wouldn't make it sit on the roof of a car. If this were the case we wouldn't be able to do a loop on a roller coaster
@@themathandphysicstutor Like without something that fastens you in to your seat like a lap bar, you would fall out of the Rollercoaster. I'm still not sure that I understand the explanation as to why the FBD is toward the track because centripetal force is toward the center.
you are confusing where the ball wants to go and where the tension force is acting. The tension is toward the center! That is fc. The ball wants to fly away but the tension keeps it in place. As far as seat belts. If your going fast enough you dont need seat belts. I guess your teacher didnt spin water over is head. There are no seat belts for skate boards who ride a loop. They stay on because of a Fc
@@kevincorrigan7784 The restraint bar on the roller coaster is only there for safety, in case the real method of holding you in your seat fails. It is not meant to carry a load under normal circumstances. In the event that the roller coaster doesn't get enough speed at the top of the loop, that is when the restraint would need to keep you in place. FYI, an upside-down roller coaster will have a restraint harness, rather than just a restraint bar like right-side-up only roller coasters have, but the principle is still the same that it is only there as a backup method of supporting you. The ride would be very uncomfortable for passengers, if it was actually built to load the restraint harnesses in tension when upside down.
What is stopping the ball from falling, such that the string is still in tension, is the fact that its immediate environment is accelerating downward at a rate greater than gravity. This means as an occupant of the roller coaster, you feel like you are pulled toward your seat by an apparent force to the outside of the track. What is really happening, is that you are accelerating downward, and it takes an inward force of gravity plus the normal force to accelerate you downward. Or gravity plus tension in the case of the hanging ball from the ceiling of the roller coaster.
For part c, wouldn't the net force in the x direction equal 0, so mgsin(theta)=Tx, so the answer would be 1sin(30)Newtons and not 1N. Wouldn't the net force in the y direction be mgcos(30), so 0.87N?
You are supposed to be calculating the tension in the string, not the net force. We are also using the coordinate system of a stationary observer, rather than the coordinate system in the axes of the car's interior. In part C, all forces add up to zero, because it is moving at a constant speed in a straight line (i.e. constant velocity). This simply means the tension is equal and opposite gravity, in order for forces to add up to zero.
The only way there can be an x-component of tension is if there is a horizontal acceleration. The only way the vertical component of tension can be anything other than equal and opposite to the force of gravity, is if there is a vertical acceleration. There is no acceleration in this part of the problem.
Hey Mr. Finn, I was wondering why for part d, you couldn't put Fg into the equation as the negative direction and Ft(vertical) as the positive direction. Also for part e, I am confused on why the ball hangs upwards and not downwards.
You could do that. But then you would have to call the acceleration negative because it is in the downward direction. I called Fg positive so I could just positive 5 for the acceleration and also us positive 10m/s^2. There was a lot of "downs"... So I called down positive that's all.
The ball is upward because and object in circular motion wants to travel tangent to the circle. So every instant it wants to fly away but is kept in motion by the string.
@@themathandphysicstutor oh ok thank you. that helps.
Hi! For the ap physics exam, do we always use 10m/s^2 for gravity?
The ref table says 9.8. Usually questions will tell you to just use 10. I don’t think this will matter. I don’t see there being much math on the exam
Great vid! Just had a question. For part C, how is there a sum of the forces in the Y direction? Wouldn't the force of gravity and the vertical tension component cancel each other out?
First, There is always a "Sum of forces". Sometimes that sum is equal to zero. Sometimes it's not. In this cause we know some of the acceleration is in the downward direction. Acceleration is caused by and Fnet. So if there is some acceleration downward, then there has to be some Fnet downward. That Fnet is the difference between Fg and Tv
@@themathandphysicstutor Thanks for clearing that up!
Super helpful video!!! Great content!
Thank You Mandi! Glad it helped!
In question d, wouldn’t the sum of the forces in the y direction be equal to Tv-mg instead of the other way around?
Mg is bigger than Tv so no
Correct Snehal
how do you know mg is winning, doesn't tv win because it is still holding up the ball?
Very helpful, thanks!!
You're welcome!
I thought for part D on inclines, sin() is always the horizontal and cos() is the vertical. Like for a box on an incline sliding down. Why in this case is cos() for the horizontal and sin() for the vertical?
Who ever told you always was fibbed to you. You have to do the Trig of the vectors to find the resultant. Try and make a little "assumptions" in physics as possible. Always solve when you can
@@themathandphysicstutor Got it, thanks.
I was taught this too, this is for the components of mass for an object sliding down an incline.
@@brianlau6273 Whether sine or cosine is applied to the horizontal, depends on the specifics of the problem, and how you opted to define your angle, or how you were given your angle. It isn't always going to be the same. A good way to determine which trig function should be associated with which direction, is to look at the limiting cases of the angles. Suppose we set the angle to zero, and suppose we set the angle to 90 degrees.
At an angle of zero, sine is also zero, while cosine is 1. The vector will align with the component axes at an angle of zero, causing one component to diminish to zero (the one that gets determined through sine), and the other component to become the full vector (the one that gets determined through cosine).
At an angle of 90 degrees, cosine is zero, while sine is 1. The exact opposite should happen for the vector's projection onto the two coordinate axes.
Wait what if we are typing the exam, how would we draw a diagram?
go to the college board website and take the demo test!
@@themathandphysicstutor ok thanks!
@@aryaanduggal1299 did u find out how?
@@spamspam3298 they wouldn't test you on drawing diagram or on complicated calculations
@@spamspam3298 yeah i did, using google docs
Lol this is very last minute but for part d why did you put that fc=(mv^2)/r ??? I thought the equation for fc was just (v^2)/ r ?
You are confusing force with acceleration. Centripetal acceleration ac = v^2/r. The net force needed to cause centripetal acceleration is Fc = m*v^2/r
Hey there, for part c, why do you substact the .25 from the 1 instead of adding it.
Not sure what your talking about. Can you time stamp the part you’re talking about. Part c had pretty cut and dry answers
in part d, i thought the acceleration was only horizontal, just like in point b, so the vertical would still be one N why is this not it?
In part b it’s on a horizontal surface. In part d it’s going down an incline. That’s the major difference
@@themathandphysicstutor got it thanks! but i still dont understand how you knew to draw a triangle in part d and label that fnet. That hypotnesuse is in a diagonal direction
I first drew the vertical and horizontal components. Then drew the direction of the Fnet which is in the direction of the acceleration. With that the triangle appeared just like in a projectile motion problem
@@themathandphysicstutor thank you! I understand now, keep making these great videos!
How are they going to expect us to do sin and cos if they don't expect us to have a calculator
All the instructions on what will be required of you is on the college board website
There are special angles with easy to remember values of sine and cosine, that you are usually either expected to memorize or know how to derive, at least in a math class. These are the ones based on the 30-60-90 right triangle and the 45-45-90 right triangle. For sine, remember sqrt(n)/2 is the value for 30 deg, 45 deg, and 60 deg, where n = 1, 2, and 3 respectively. Obviously, sqrt(1) reduces to 1, so that trig value is 1/2. For cosine, just follow the same pattern backwards from 90 degrees to zero. Tangent would come from sine/cosine. Additionally, you should know what the values of sine and cosine are at integer multiples of 90 degrees, based on the general pattern of the waveform.
I'm not sure whether they would expect the same degree of memorization on a Physics exam. If it were an exam I were to write, and for some reason that was not up to me, no calculators were permitted, then I'd allow students to receive full credit for writing sin(60 deg), just as a student would get full credit for knowing this was equal to sqrt(3)/2. According to the college board's website, they do allow scientific calculators which would have the ability to calculate sine and cosine.