Pulley Definition and 1000 Threads

##Net force=50-T2+T2-T1+T1-20=10a## ##a=3 m/s^2## I want to ask when we do this way rather than taking individual masses we can’t decide the direction of the system as we can when they are taken individually. So is it correct to just leave ##a## as it is and solve ?

I am having trouble creating the constraint equation for this pulley system. I don't understand why the last 3 variables of the following constraint is divided by 2? Could anyone help me understand why this is?

1) By the Work-Energy Theorem, ##W=K_f-K_i=\frac{1}{2}I_{0}\omega^2=\frac{L^2}{2I_0}.## 2) By assuming that the initial length of the spring is ##0##, calling its final length ##S## and ##T## the tension in the rope connecting the pulley and mass ##m_p## I have: ##\begin{cases}(kS-T)r=0\\ m_p...

1. ##-f_k\cos\theta-T\cos\theta+F_n\cos\alpha=m_2a_x## 2. ##f_k\sin\theta+T\sin\theta+F_n\sin\alpha-m_2g=-m_2a_y## 3. ##T-m_1g=m_1a_y## I am unable to get anywhere. There are accelerations in x , y directions. I need the value of acceleration. Then I can simply use ##s=ut+\frac12at^2##

We can differentiate twice the y displacement with respect to time t and get the acceleration of block B. $$a_B= \frac12 m/s^2$$. But I don’t think it’s that simple.

I was researching a problem that had once been posted here by someone else, and that had subsequently been posted (couched in somewhat different terms) here by me, and in doing the research, I ran across the following, which I think is the well-stated 'original version' of the problem. The...

I have no clue how to do the problem. I created a free body diagram for each block. I assume that it is simpler than I am making it out to be.

##(T-m1g)-(T-m2g)=(m1+m2)a## But I think this is wrong eqn.

I want to ask in this machine when there are two masses hanging down then the tension T is directed upwards along the rope. Is it the force applied by the rope on the mass? Is it the force applied by the pulley? When the anyone mass of the machine moves downwards is it because the force of...

##(T- m1g)-(T-m2g)= ( m)a##. I don’t know what m?

This problem just came to my mind when thinking on another problem. Does the tension is just 2T as it is if the angle "a" is 90 degrees? It seems not to me. In a "normal"( I don't really know what is the right word for that) situation, the tension is would be 2T at the line in the middle of two...

I am making a system Where I am pulling on opposite ends of a pulley with Tensions T1 and T2. Is this equivalent to preloading the pulley with tension T2? If I preload the pulley with T2 on a separate rope, how will the angular velocity change when I apply T1? I've drawn a quick diagram

To find the tension in the rope connecting 6.0 kg block and 4.0 kg block we do 6.0 kg = m1, 4.0 kg = m2, 9.0 kg = M (m_2 + m_1)a - Ma = Mg - m_2 gsin\theta - m_1 gsin\theta Why do we use sin in these equations and not cos?

So far I have: The velocity of the belt will be the same for pully A and D, so we can calculate the angular velocity of pulley D: ## V_A = V_B ## ## \omega_A r_A = \omega_D r_D ## ## ((20*3)+40)(0.075) = \omega_D (0.025) ## ## \omega_D = 300 Rad/s ## My next step was to determine the angular...

> The set up: At left end, the rope is pulled down with a distance $dx_1$ by a force of constant magnitude $F$, the mass of $M$ is wrapping around by rope on the right and moves up by a distance $dx_2$ due to this. Problem: Find relate the two displacements. I thought of applying energy...

First I calculated the forces that were working against mass B. m(A)g sin 30 + μm(A)g cos 30 = 12.86 N The force working with mass B is m(B)g = 9.8m(B) I thought I could solve for B using F=ma where 12.86 N = (2kg+m(B))*(0.58), but of course, 12.86 N is just the force required to make the...

As the force on a pulley is equal to twice the tension, I just have to find the tension to solve part A. To do so, I first wrote the equations for both m1 and m2. m1 * a = T - m1g m2 * a = T + N - m2g The tension must have the same values for both equations so I added both equations to find...

This is not homework. I found this question when browsing and there is also solution but I do not understand it. This is part of the solution: My questions: 1) From equation (i): ΔT = m.a, I think ΔT refers to W - T (weight of block - tension of string). I got this from free body diagram and...

My attempt was to calculate the acceleration of M2 as the acceleration of M2 if it were the only mass in the system, minus the component of M1's acceleration along the slope. And then I would divide the whole thing by 2 to get the acceleration for just one of the two masses@ a = 1/2 ( g -...

Hi everybody, first message on this forum ! I am building a machine for a friend who works with cardboard pulp, he is also a performance artist and would like a machine where 2 or 3 performers can activate the machine by hand with pulleys and wheels, and shred the pieces of cardboard that are...

I've been stuck at this pulley system for a while. I've calculated the force of which A pulls => FA = sin25*50*9.82 = 208.5 N But I get stuck on the free body diagram. Can someone help and explain the freebody diagram

Below is an ideal mass pulley system that we encounter in many problems under Newton's Second Law of Motion questions. Its supposed to be massless and frictionless i.e. string slips over the pulley and pulley does not rotate. In a real system, the pulley is assumed to be massless, whereas in...

Hello, I'm working with a pully system and can't find the answer for this question on Google. I'm wondering how to calculate the load on the rope in order to determine what grade rope I should use. For example let's say I'm picking up a 100 lb box using a 4 to 1 system. Not accounting for...

i) I first analyzed the forces as soon as the 14 Kg is released. The aim of this step is to calculate the work done by the net force acting on the 14 kg mass to determine the change in kinetic energy. T-14g=-14a T-8g=8a T=99.7 N a=6g/22 m/s^2 Since the net force is constant and does not vary...

I assumed that each of the masses attached to the string would be pulled with a force F, instead of F/2, which appears to be the correct value. Why is it F/2? I suspect the fact that the pulley is "weightless" has something to do with the F/2 value. What is it? Thank you.

Summary:: I am designing a linear telescopic system that gets elevated with a step motor, but in order to find the compatible motor, I need to work out the torque that is required for the elevation. I have uploaded my CAD model to this thread with how cables are connected. Could someone please...

1 Block m2: m2g - 2T = m2a2 (1) Block m1: ∑F= ma T= m1a1 = m1.2a2 (2) Cancel T in (1) and (2) I have : m2g = a2 (4m1 + m2) => a2 = m2g / (4m1 + m2) => a1 = 2m2g / (4m1 + m2) Is my answer correct ? Thank you

Summary:: Please see the picture below Let say: ##W_1## is weight of ##m_1## ##W_2## is weight of ##m_2## ##f_1## is friction on ##m_1## ##f_2## is friction on ##m_2## I want to find the acceleration of the system. Since I don't know in which direction they will move, I just assume ##m_1##...

I have a question about friction between pulley and rope? Does anyone know how to calculate friction between pulley and rope I enclosed an example picture Thank you!

There are 3 different strings in this system. The one pulling P, the one pulling F and the one pulling a pulley. Since the questions says they're ferried across before hitting mast or deck, I'm assuming that they are not stationary. g is gravitational acceleration. P will move twice as fast...

I did part a and part b but stuck in part c. Could you help me? (Part a and b attached.)

This looks like a classical setup but I can't find a solution. We can calculate the energy of the system by looking at the work done by the gravity and the spring. But how do we divide the energy between the kinetic energy of the pulley and the rotation of the pulley?

I tried hard but was not able to make progress. Problem no 33.

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 have...

I approached this question with a free body diagram. my Verticle length of BA is y-2 AC= 15-AB Horizontal length of AC= a; Horizontal length of BA=10-a I have too many unknowns in this problem and I don't know any directions(angle). I know that I should start at the weight and try to resolve the...

I solved for T on m1 and arrived at 6.72. I plugged that value into the ΣFx equation as shown above (pardon my handwriting) and got a mass of 0.88 kg. The online program indicated that I needed to check my expression for tension, noting that the two tensions are heading in opposite directions...

From this question, I do not understand why there are three forces exerted at Point C (2 of it being the tension by weight A and the other is the tension by weight B) I understand that there is tension by the two weights but why is there 2 forces exerted by weight A at point C? From the...

Fsp = 90 x 0.12 = 10.8 Ffriction = Magcos(titre) x 0.30 I got the answer 2.09ms2 when the correct answer is 1.11ms2. What am i doing wrong here?

The normal reaction from the ground and from block m, the force of gravity are the forces I feel should act on the wedge but since the wedge is a perfectly rigid body and the pulley(which is massless) is attached to it so will tension also act on the wedge as well?

FBD Block 1 FBD Block 2 FBD Pulley B I'm mainly concerned with the coordinate system direction in this problem, but just to show my attempt, here are the equations I got from the system. ##-T_A + m_1g = m_1a_1## ##T_B - m_2g = m_2a_2## ##T_A - 2T_B = 0## Using the fact that the lengths...

Hello everyone, this is a thought experiment I made, it involves two ropes being pulled of a pulley similar to an atwood machine, and a block attached to the two pulleys at the other end. The ropes are being pulled at a constant velocity $ U$ from the end away from the block and the block rises...

If a free body diagram is constructed for 1kg, one might be inclined to draw an arrow downwards with only the weight of 10N. However, the downward, non-net, force on 1 kg is (1+5) x 9.8. Why is this the case? It seems very counterintuitive.

. First thing that I notice is that the other part of string (to which body 2 is connected) is fixed. Therefore, I concluded that the body 2 is stationary with respect to the pulley 2 (let's agree that pulley to which body 1 is connected is pulley 1 and in same way pulley 2 is defined)...

If M moves ##x## along the plane, her height variation in ##x \cos(\alpha)##, and, but I don't know how to find the variation of the height of ##m##

I remember, a few weeks ago, when looking for homework problems I could help with, seeing one as described in the title. I couldn't think of an easy solution and was busy at the time so i made a mental note to think about it later. I finally got around to doing that yesterday, and brushed up on...

Hi, I am looking for some guidance on how to approach this calculation. I have an air cylinder operating a lever assembly that then applies pressure to a pulley of which a belt is wrapped around. I need the belt to have about 4500 lbs of tension. How do I work backwards to figure the required...