Pulley Physics Problem - Finding Acceleration and Tension Force
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- เผยแพร่เมื่อ 2 ต.ค. 2024
- This physics video tutorial explains how to calculate the acceleration of a pulley system with two masses with and without kinetic friction. It also discusses how determine the tension in the rope as well. The formulas and equations are all provided in the two practice problems & examples presented in this tutorial. The full version of this video also explains how to solve the pulley problem with two hanging masses as well as how to determine the direction in which the system will move. It also explains how to solve pulley problems with inclined planes.
Full 1 Hour 28 Minute Video:
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PDF Worksheet - 8 Problems:
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Forces - Formula Sheet:
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Full 1 Hour 28 Minute Video:
• Pulley Physics Problem...
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Direct Link to The Full Video: bit.ly/3QdBavj
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Thank you!! Pulleys are so much easier with Sum of forces over mass.
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Hi, there's something I'm confused about. When considering the rotational inertia of the pulley and doing a problem with it, it becomes apparent that each side of the rope has its own unique tension that is not equal to each other. In this case, doesn't the method to solve problems without the consideration of pulleys violate this principle? What assumptions are made in order to solve it like this? If it's the existence of the pulley itself that causes the uneven tension, what exactly is the pulley doing to the string to essentially split apart its effects?
(the assumption made, by the rope being massless, is that tension is evenly distributed in the rope)
Frictionless pulley; massless and perfectly inelastic rope
Massless pulleys to be added to the assumptions
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how to find the tension of the 15kg box which is at rest ?plzz someone answer 😢
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why was gravity (9.8) positive? If the box is going downward, wouldn't it be negative?
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Great video, although, my main problem in understanding was when solving for tension. When we made the tension equation, why couldn't we plug in the value at that moment itself?
when he was solving for the acceleration he used T= to make two equations to set the whole solving part up. at that point we did not know the tension, which is what he solved for later
For those that want some more practice on pulley problems, I've uploaded a video where I solve such problems step-by-step! It might help some of you out there :) th-cam.com/video/y7g9gHchOIk/w-d-xo.html
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Good job! This helped a lot.
Organic tutor help me out this question.... An elastic string is stretched between two fixed points which are 1 meter apart. An object weighing 3.7N is suspended at the middle of the stretched elastic causing the midpoint to drop by 0.11 meters. Calculate the:
(a) angle of each part of the string to the vertical
(b) Tension in each part of the string.
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I'm a bit confused. In the 1st example isn't the weight of the 5 kg box supposed to equal to the tension experienced by the string (T=m2g)? The box itself is always attached to the string so the forces must always be constant right? Physics is not my strongest subject 😢😂
👀💀uhhh so I tried to solve the first question on my own before watching and I got twice your values still confused at what went wrong
My answers did add up though
The tension force is the same for all rope right? that’s why we can calculate it with any one of the two equations?
Yes, same as the acceleration
yes. you can choose any of the two
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so then a = (m2g - T)/m2 and a = (T - Mkm1g)/m1 also work if you have tension. That's pretty cool. But because we haven't found T, we're substituting the T instead.
Thank you so much, i had no clue how to do pulley questions, but now i understand how to analyze the question and find the answer thank you !
can anyone answer why he made ay and ax just a because they’re going different ways
what if the m1 is sliding along an inclined plane?
I still don't understand. When do we decide that the tension on the first mass gonna be equal with the tension on the second mass? At your previous video, you distinguished them both by using different indexes, which are T₁ and T₂. It's in this video th-cam.com/video/PbtEzXNnXqs/w-d-xo.html
But here, you didn't do it. You made them equal. Can you explain it please? Or anyone else that understands?
If I am not mistaken it is because since both the objects are connected by the same string the tension throughout remains the same.
The tension on the first mass is equal to the tension on the second mass because of Newton's 3rd law: for every action, there is an equal and opposite reaction. Note that the two tension forces are the forces the two masses exert on one another. The tension force at M2 is the force M1 exerts on M2 and the tension force at M1 is the force M2 exerts on M1. These forces are equal and opposite so they actually cancel out. The pulley just redirects the force of M2 at 90 degrees. If you imagine rotating everything after the pulley CCW 90 degrees, you see that the tension force on M2, M2, and M2g all lie in a straight line with respect to M1 and the tension force at M1. Also notice that now, the tension force at M2(force of M1 on M2) opposes the tension force at M1(force of M2 on M1). Because these two cancel each other out, you have an Fnet that is acting on both of these masses, that being, M2g. This tells you, M2g = a (M1+M2). Solve for a and you get, a = M2g/(M1+M2). Once you have this, it's easy to solve for T as is noted here on the video as T is the force accelerating M1 while the force accelerating M2 is equal to T-m2g. The take away you should get from this is that since the tension force in the rope is the same, but in opposite directions, they cancel out and you are left with two different masses being accelerated by Fnet(M2g). Since they both have the same acceleration (connected by a 100% static rope is the assumption), per Newton's second law, F=ma, each of these different masses must be experiencing a different Fnet given the fact they even though they have different masses, they are accelerating at the same rate. The only way this is possible is if each were accelerated by two different net forces, Fnet1(net force acting on M1) and fnet2(net force acting on M2). Note, these two net forces are not the same as the net force M2g. Remember, M2g is the net force acting on (M2+M1...imagine M2 & M1are fused together as one single large mass). It accelerates them both(imagine one large mass) and from this acceleration, we can determine the individual net forces acting on each of the masses. This is the takeaway one should get from this pulley problem. The answer is meaningless if you don't understand what's happening and why. Hope this helps!
I was so confused with the positive and negative signs, but when u talk abt it using the x and y axis, it makes sm sense!
If the thread of pulley is elastic, will acceleration change?
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Ohhh so its 2 sources of mass, but only 1 source of downwards force, m2. I get it!
Organic tutor help me out this question.... An elastic string is stretched between two fixed points which are 1 meter apart. An object weighing 3.7N is suspended at the middle of the stretched elastic causing the midpoint to drop by 0.11 meters. Calculate the:
(a) angle of each part of the string to the vertical
(b) Tension in each part of the string.
What the fuck is the 9.8 come from?
that's the force of gravity; g is always equal to 9.8m/s^2 on earth
In 5:42, how come a_x is +, a_y is +, but m2ay -? Is it just becausethe m2g vector is pointing downward, or is there another reason?
hold on the net force is going downwards, shouldn't it be m2g + m2ay
You sir are a god when it comes to explaining things.
Pardon me. 14:44 How the Fk=(constant) m1(g). How is that the force is accompanied with g? Isn't that it is some kind of (a)?
Bro just take g=10m/s^2 when you are teaching.
But what if you have to find the kinetic coefficient and you have to find the acceleration but the only thing you have is velocity, mass, and time?
Why label ay as negative? If you label Tx as negative, given that the force in x direction, Tx is therefore negative, wouldn't that give you the right net acceleration, without having to artificially add a negative sign to the ay?
It’s based on a coordinate plane. To the right on the x-axis is positive, to the left on the x-axis is negative, upwards is positive on the y-axis, and downwards is negative on the y-axis. Since block m2 moves downwards, it is negative. Hope this helps!
@@fridamoreno9689 Hey, just realized that he labels ay as negative because in his coordinate plane, downward=ay and upward=-ay. Therefore, he says ax=ay since m1 follows a path to the right and then downward. But if upward=ay and downward=-ay, one would have to make the substitution ax=-ay, since m1 follows path to the right and then downward.
Why can't we take 'g' as negative if we have taken 'a' negative in the negative y-axis?
How do you get the 36.75N? comment please
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Dear sir,you are my hero,my savior and the best teacher. Thank you for this explanation 😊 I'm am really grateful to you😭😌❤️
Sir I. from watching your video? now I have questions what was the tension in the system and what was the purpose of driving that equation?
What’s up with all the variables though? Like instead of m2g can’t you just say F2 so it’s easier to understand?
What would happen if the sliding block, is on an incline with the ramp pointing towards the pulley ?
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how can you find the mass of the hanging mass??
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for the first problem, why is the horizontal tension equal to the vertical tension in the string?
By seeing just first 2mins of the video u will understand how to find acceleration
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Wow, this solved all the confusion I had when trying to find Tension.
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An 80 N object suspended from a ceiling is pulled by a horizontal force of F, as shown in the figure. a) Find the tension in the string.
Who can help me pleasee?
isn't net force always addition so for the pulley problem around 3:29, net force for Y should equal T + m2g?
No because they go in different directions T goes up and m2g goes down for exemple let's say you took 3 steps to the right and then 2 steps to the left the final spot from the start is 1 steps to the right because 3-2=1 so it should be T-m2g. Dont know if I explained it well hope you understood
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important stuff is spoken unclear and non important stuff is draggy...
Ok how do you find the mass of object 2 if given the acceleration of object 1
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Please can u teach only the derivation of the equations and the their concept.
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