Pulley Definition and 1000 Threads

Dear all I have the following problem with my working attached as a PDF. I want to calculate how far particle a travels after the string has snapped. The question says the particle A travel for 1m before the string snaps. The question asks me to use energy methods to do this. The total distance...

The following system is at rest with a block of mass m1 = 5kg, and a force F that is applied to the end of a string attached to the pulley system. Let g, the acceleration due to gravity equal to 10 (for simple calculations). My analysis is as follows: I'll first label the strings Tsub1 and...

Building a DIY kayak simulator training machine. A rope is tied to each end of a broomstick and runs around a resistance wheel, guided by a system of seven pulleys . The machine works exactly as intended. But I feel there is too much drag. Trying to see how I can reduce the drag,in the...

I suppose to keep the pulley from rotating the net torque has to be zero? Let ## F_{r} ## be the force that the 5 kg mass on the ramp exerts on the pulley and ## F_{d} ## be the force exerted straight down by the other mass on the pulley. Let ## r = 0.3 ## m be the outer radius of the pulley...

Hi. The following situation: The pulleys are fixed to the floor/ceiling and massless, as are the ropes, and there is no friction. At ##t=0##, the masses are released from rest. For the moment, I'll assume ##m_2\gg m_1##. So ##m_2## will accelerate at ##g## and the red rope at...

TL;DR Summary: Need ideas on how to get maximum weight/resistance out of a pulley system in a centered axis extension machine Hi so I am studying mechanical engineering and am currently working on a factory that makes gym machines, and I wanted to know if anyone could explain to me how I could...

Hello! I have this problem from an old exam I'm trying to solve. The problem is in Swedish so I've translated it: NOTE that I accidentaly wrote $$C\neq 1$$ in the picture below. The correct problem statement is above. But that part is not what I have problems with. The answer key says "if the...

The question is regarding this image, I understand the two 50N forces provide a clockwise moment of 40 x 10^-2 x 100 which gives option D, as this moment must be balanced by the moment given by T. Now my question is, how is the net force on this system equal to 0? It says it is in equilibrium...

A kid is standing on a scale and holding a rope that goes through a pulley, on the other side the rope is connected to a spring and then to a 50N mass item. Assume that the kid's weight is 700N 1. While in static mode - what will the scale show? 2. The kid is now pulling the rope to get the mass...

many pulleys at various angles multiply the initial power of the winch

Please note that the system is not in equilibrium, and that tension must be solved for the instant where the angle is 154 degrees. My attempt (correct ans is Ft = 626N)

Dear readers, I want to design a pulley system like in the schematic below, a system where the distance between certain attachment points are equally spaced along a vertical bar. The main feature of these pulleys is that when a force (F) is applied on one end of the cable, the attachment...

For A the 1.2 kg block is being pulled by gravity hence work is done downwards which will make work positive since it's going with the same direction as the force. 1.2 * 9.8 = 11.76 N pulled downwards Work = F*d 11.76*0.75 = 8.82 J The tension is the other force and since the thing is...

The elastic force of the board,N=G_1+G_2=1000N. The kinetic friction f=μN=200N. The tension T=f=200N.

My question is how come the tension in pulley 2 is pointing downwards. I was under the impression that tension always points away from the mass.

15 - Tension = 1.53*a T = 15 - 1.53*R*α T*0.33 - 1.10 = Iα Is this approach correct?

Hi, I have this pulley question, i have to find the force F that needs to be applied to the string to keep the system in equilibrium. I found 150N. Is it correct? T_1= 300N and T_2=150N so F is 150N.

Set up the force equations: (1) ##mgsin(\beta) - T = ma_{1}## (2) ##TR = I_{s}\alpha_{1}## Multiply (1) by ##R## and isolate ##TR##: ##R(mgsin(\beta) - T) = R(ma_{1})## ##mRgsin(\beta) - TR = mRa_{1}## ##TR = mRgsin(\beta) - mRa_{1}## Plug ##TR## into (2): ##TR = I_{s}\alpha_{1}##...

What is the initial acceleration of mass 5M .The pulleys are ideal and the string inextensible. My attempt- 2Mg-T=2Ma (for 2M) T=Ma (for M) Solving we get T=2Mg/3 T-N=5MA (for 5M) N=2MA (for 2M) Solving we get A=2g/21 but the given ans. is 2g/23

This was the question: I derived the equations as mentioned in the relevant equations. But I could not solve the equations to find the answer. I realise with respect to inclined plane the acceleration must be a since string cannot slack. With respect to ground, the acceleration of incline is...

I started by summing the forces and torques to get: - ma = mg-T - I*alpha=Tr I then used a=alpha*r and I=Mr^2 to combine the equations and solved for angular acceleration equals 81.75rad/s^2. Plugging this back into a torque equation I got that the net torque is 1.04Nm. However, the problem...

I tried to multiply 1/8 g (1.22625) by the radius (1.25 m) and got 1.53 rad/s^2. This is actually the linear acceleration of the elevator. How do I get the angular acceleration of the disk? Thanks!

For this problem, The solution is, However, how do they know the block B will move up and block A will move down? The masses of each are not given so could be the other way round if ##m_b > m_a##? Also how do they know that if block B moves up by 1cm, block A will move down by 2cm? Many thanks!

For this problem, Why is the tension on each side not equal? For this problem I think the only assumption is that the string is inextensible so the accelerations of the masses are equal. Many thanks!

First, I assumed that the tension in the rope connected to the block A equals the static friction ##\sum{}^{} F_x =0 \rightarrow T=N_A*0.5=100N##, then the W weight or force equals to the tension in the pulley and the tension T ##W=100+\frac{2W}{\pi} \rightarrow W=275.2N## It's the first problem...

^ This is my personal drawing of the diagram, I couldn't take a picture of the actual one. The setup is a pulley wrapped with a cord and mass hangers attached to each end. My first thought when approaching this problem was to first determine the rotational inertia of the pulley, then use some...

So far I have used methods of joints to determine the forces at point E, D, and C. However, there is also a pulley attached to point D and E which I included in the sum of forces, but I'm not sure if that's the correct way to apply them. Every force I've calculated so far has also been in...

in the opposite problem that has three pulleys and two masses, the book is asking for : 1- the tension on the string. 2- the acceleration of pulley p1. 3- the acceleration of mass m2. 4- what should the mass "m" be such that m1 does not accelerate? note : pulley p1 has mass m, in the diagram of...

I have a VPB-808 Fan (rated 4100 RPM) from Viron. I am currently trying to pull 1300 CFM at 7" SP, which needs ~2668 RPM (2.85 HP). I have a pulley drive motor (Baldor, EM3611T, 3HP, rated 1760 RPM, 4.2A). It is drawing about 2.8A with the following configurations. It is currently only blowing...

I have a conceptual question about this problem. I can write the 3.5 kg block equation as Fnet(block 1)=(Force of tension)-(Force of friction)=m1a I can write the 2.8 kg block as Fnet(block 2)=(Force of tension)-(Force of gravity2)=m2a My question is this If I set the forces of...

I got 13N but is that right because apparently, it's wrong Here's my work: F = mg = 2(10) = 20N F = ma a = F/m = 20/4+2 = 20/6 = 10/3 = 3.3m/s^2 T = mg - ma T = (2kg)(10m/s^2) - (2kg)(3.3m/s^2) = 13.4 N I appreciate it! And if I'm wrong could you show how you got your answer? Thanks

I know that the tension from pulley B (T1) has to be equal the m*g of m1 for m1 to have acceleration = 0. But i can't figure how this works because the m2 is already heavier. And so the block(M) has to be negative weight?

I used m1=m2=15,0kg System: FBD: Note: I believe I have solved a) correctly and am more confused about b). a) I started with drawing the FBD. Knowing that the net sum has to be zero for the system to be at rest I used the left mass to find the tension on the rope: S1 = G1 = 15kg*9,81m/s2 =...

I'm attempting to repair my Ergotron Workfit station. After taking apart, I discovered the problem. A single nylon rope split, releasing the spring's tension. Because it is past the warranty, the manufacturer is recommending I toss the whole assembly and buy a new $650 one. Doesn't seem the...

Hello there, can I get some help with (b) please? first of all I wanted to ask.. can is it permitted to use different systems in one exercises? like f.e. for conservation of energy to find the velocity of ##m_3## I used as system only ##m_3## but for the collision I used the 3 masses as the...

Is my answer correct? And if not, why?

I'm having some kind of mental block. If I extend ##l_x## by ##\delta## ,I expect the hanging mass to move ##\frac{ \delta}{2}##. I can't figure out how this is the case from the constraint: $$ l_x+l_1=C $$ ##C## is an arbitrary length I keep getting that ##l_1## changes by ##\delta##, but...

There is no friction mentioned by the question so I assume the plane is frictionless but can the sphere roll without slipping if there is no friction? This is my attempt: Equation of translation motion of object A (assuming A moves upwards): TA - WA sin θ = mA . aCOM (A) TA = mA . aCOM (A) + WA...

(a) By setting up a coordinate system with the x-axis pointing to the right and the y-axis pointing downward we have ##\begin{cases}-kx_{eq}+T_1+F_{s}=0\\ -RF_{s}+rT_1=0\\ r_p (T_2-T_1)=0\\ -T_2+mg=0\end{cases}\Rightarrow x_{eq}=\frac{mg}{k}\left(1+\frac{r}{R}\right)## which coincides with the...

Below is a pulley with mass and a string around it where the tension of one end of the string is 20 N and the tension of the other end of the string is 60 N. I know there is a net torque due to the differing tensions of the string, so I am wondering what the translational forces are on the...

Setting up coordinates for the problem ##L=\frac{1}{2}M_1 \dot{x}^2+\frac{1}{2}M_2(\dot y-\dot x)^2+M_1gx+M_2g(l_a-x+y)## After using Euler Lagrange for x component and y component separate and substitute one to another then I get that ##\ddot{x}=\frac{M_1-2M_2}{M_1}g## whereas on the...

A) I am not sure if initial velocity means the initial velocity of the putty. If then, it should be 10m/s. But the answer given is 10/7 m/s B) i have solved this part correctly C) the answer should be 86J Please be kind to help me

Hello all, this is an adaptation of a question i saw some time ago (can't find the original now). There are two forces acting on two masses both 1kg. The masses are joined by an inextendable rope and going over a frictionless pulley of negigible mass. In blue I have written in some working out...

torque = rF = I*aT/r F = 0.5*aT*m_p W = F (tension in rope)*d = 0.5*aT*m_p*d Don't know because of unknown d

##m_{A} = 3 kg## ##m_{B} = 2 kg## ##y_{A} + y_{B} = c \Leftrightarrow y_{A} = c - y_{B}##, where c is a constant. ##\Rightarrow \dot{y_{A}} = -\dot{y_{B}}## The Lagrangian: $$L = T - V$$ ##T =\frac{1}{2}m_{A}\dot{y_{B}}^{2} + \frac{1}{2}m_{B}\dot{y_{B}}^{2}## ##V = m_{A}g(c - y_{B}) +...

A leg is suspended in a traction system, as shown in the attached image. (a) Which pulley in the figure is used to calculate the force exerted on the foot? (b) What is the tension in the rope? Here T is the tension, wleg is the weight of the leg, and w is the weight of the load that provides the...

Consider the pulley in the attached image to be frictionless. (a) If m2 is released, what will its acceleration be? My question is why wouldn't m2's acceleration be greater if released rather than attached to the string because m2 released is no longer subject to the string's upward force tension?