Thank you for the response. This answers my question.
I think my misconception came from thinking that the object (more dense than water) would have to be sufficiently heavy to displace more water when thrown overboard.
However, my working shows that, under the assumed circumstances, m could...
I was asked this question:
Assume you're sitting in a boat (you and the boat, together, are a mass M) which also contains a heavy object (of mass m). The boat (inc. you and the heavy object) is floating in a pond (a fixed body of water, rather than open sea). If you throw the object overboard...
Thank you for answering my questions.
I will look more into 'force due to pressure by integration'. I'm not at university yet, so I haven't had to use integration in real-world problems, but it could be useful to learn in-advance.
Is the horizontal component of the area the same as taking a horizontal cross-section of the shape up to where the fluid line reaches?
For example, if you put in a cone with a circular base into fluid (tip-first - i.e. with the tip facing downwards), would the horizontal area be the same as...
I think I understand now. As long as there is fluid in-contact with the base surface of the object, there will be an upthrust.
Does this upthrust always equal the weight of fluid which the object displaced, regardless of what percentage of the object is submerged?
I'm assuming that a base...
I had some questions about Archimedes' Principle. This image shows an object being gradually submerged into fluid:
As far as I'm aware, when the object is floating, weight of fluid displaced by object = upthrust acting upwards on the object.
When the object is fully submerged, volume of the...
Why do we use the initial velocities of zero when you could use the velocity immediately before the collision (i.e. X has velocity v)? How do you know which to use? I understand the momentum calculation, but I don't understand why you don't use the velocities right before the collision.
I don't understand why.
You have the angle θ/2 in the bottom-right corner (between ramp and string). The 90 degree angle is at the top corner. The triangle appears to also be isoceles.
If you use the angle by the string, you get x2 to be the opposite side and the ramp to be the hypotenuse...
I don't understand the reasoning of this question's answer. The answer is velocity = 0 (option A).
A while ago, I was told that, since the magnets were held at-rest (before being let go), they must have no velocity after the collision. What about the velocity which they had just before the...
I can't see where I've made a mistake.
I just used: moment = force x perpendicular distance (from hinge to direction of force)
If you make a triangle connecting hinge and force going down from centre of ramp:
sin(θ) = O/H
sin(θ) = x1/(l/2)
x1 = lsin(θ)/2
CW moment = Wlsin(θ)/2
If you...
I did this question a while ago, but I wanted to ask how you know that the maximum tension occurs when θ = 90 degrees.
I worked out the moments, and I got:
tension = Wcos(θ/2)
I'm thinking that I did it wrong.
We want cos(θ/2) to be as large as possible to maximise tension. Of course, θ...
Thank you. I now know I should have given F a '-' sign in-front, since force is acting in the negative direction, as I assumed positive to be right to left.
initial velocity = +u
final velocity = -v
F = change in p/change in t
-F = (-mv - mu)/(t2 - t1)
-F(t2 - t1) = -mv - mu
mv = F(t2 - t1)...
Did I answer my own question in the final paragraph of my post? If so, I think that's all I needed to know for this problem. I believe it makes sense to me now.
Please scroll-sown to Question 52: https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf
The correct answer is 'B'. This is the working I did:
F = (change in momentum) / (change in time)
change in momentum = mv - mu, where v = final velocity and u = initial...
Please scroll-down to the end (Question 54): https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf
I have also been referring to unofficial worked solutions (http://www.engineeringadmissionsassessment.com/2017-solutions.html), but I didn't understand how it...
Please scroll-down to Q50: https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf
The correct answer is 'B', or 'mgsin(Θ)'. I put 'E', or 'μmgcos(Θ)'.
There are unofficial worked solutions which I have been referring to when I have attempted the question and...
I was taught to write the equations of motion such that the directions moved by each object would be consistent (e.g. if one goes up, the other goes down). If an acceleration comes out negative for an object, it means that the direction of acceleration you predicted was actually opposite for...
I think I get what you're saying, but I'd like to clarify.
If you make the decision that both masses go down, do you then decide to change the sign in front of '2aB', since you know that, in reality, one will go up if the other goes down?
Is the reason for using 'aA = +2aB' when one mass goes...
I managed to do the first pulley question, though I have a question about this second one you have mentioned.
Why do the equations of motion for masses A and B assume the acceleration direction to be down? Why does the answer come out wrong if you assume one mass goes up (for A: T - mg = maA)...
In post 21, what is this 'superposition principle' you mentioned? How do you know that the 'acceleration of the small pulley relative to big pulley' + 'acceleration of block relative to the small pulley' = 'absolute acceleration of the block'?
Oh, I assumed ma = 0, since we are told the pulleys are light (negligible mass). If using forces is wrong, I guess it doesn't matter now that it's wrong.
I don't really understand how to solve the problem with kinematics. This problem is a mess.
I have attempted the original question from the beginning. It spans multiple pages of solving simultaneous equations, so this post will only show the equations without their derivation. This is what each variable seen in the equations refers to:
T1 = tension in string connecting block A and...
We only know B and C are not 'accelerating'. That doesn't mean they're not moving position.
If the left-pulley rotates, one mass has got to go up and the other has got to move down, correct, since the string is being pulled down more to the left/right by the heavier mass?
I don't see how the...
I just think that I'm nothing exceptional, regardless of what it may be. Then, I ask myself, if I don't have any special skills, why would the world benefit from me studying engineering, in order to develop technology that makes lives better? There are many people on this forum alone that could...
I see what you mean by the pulley rotating about its centre, since it has the weight forces of both masses acting on opposite ends of its circumference.
If A has a=+4 (going upwards), does that mean B has to fall, therefore meaning that mass B is heavier than mass A, and the pulley is rotating...
If B were stationary, wouldn't we have to assume that A wouldn't move either? In the actual question, if A has a=4, why doesn't B have a=-4, if they are connected particles? I have the same confusion with C and D. I don't understand how this system works.
Sorry, I don't understand what you mean at all. Why does movement of the pulley affect movement of the mass blocks?
I am beginning to think that my application to university next year was a terrible mistake.
I don't understand the working shown on the website (seen above).
Firstly, they establish 'up' as the positive direction for acceleration, so A is +4 and D is -4.
They create two points, P and Q, and set acceleration values for them. I understand that connected particles have the same...
In the original problem, why do we know that the pulley is rotating about its diameter, as opposed to its rotational axis?
Was the reason I gave right for explaining the resultant force?
Is there a way to derive a shape's moment of inertia formula (does it go into more-complex mathematics that I wouldn't understand yet)? Is the video I linked to incorrect at timestamp 5:35?
As for the resultant force, is it 'T2 - T1' because T2 aids in the pulley rotating clockwise whereas T1...
Note: the working (taken from iWTSE website) refers to inertia as the symbol ‘J’ (in-case there was any confusion).
I found equations of motion for mass m and 2m which were ‘T1 = ma + mg’ and ‘T2 = 2mg – 2ma’, respectively. I know they are connected particles with the same acceleration ‘a’...
I see the mistake I made with the vector addition. Now, I can see that the triangle is equilateral. I understand what you said about the mass lying under the centre of said-triangle, and that the vertical tensions of all three strings must equal 10g.
However, I do not know how to use...
I think I've made a mistake with the vectors. I was trying to get to the same point as the original working (made by IWTSE, not me), but one of the sides of a triangle can't be 0. I would appreciate some help.
Thank you. I now understand that '2n-1' (or '2n+1') is meant to represent an odd integer, since if delta theta = 270 degrees the hands are still perpendicular. I got an answer of 44.5 (or 43.5, depending on the expression used) which rounds to 44, the correct answer.
I understand this working all the way up until the '2n-1' part, where n is a positive integer.
I understand that delta theta is 90 degrees (i.e. pi/2 radians), as the hands are at right angles to each other.. I also understand where the angle equations are derived from and why you have to find...
I think I have done this in an inefficient way. I assumed from the question that the three balances were of equal mass, though I'm not sure if I should have ignored their mass entirely. The correct answer is X = (62 + [2/9]) kg. Please help.
Just to clarify, reaction force can exist with just one (not necessarily both) out of these components: 'friction' and the 'normal reaction force'?
Also, why is it we don't know the direction of the reaction force acting on the hinge? I get why you can ignore this when solving this question...