# Problem related to pulley system and velocity

## Homework Statement In the given figure if the masses A and B move down with a velocity 'v' at any moment then what will be the velocity with which mass W rises.

A purely logical answer would be appreciated and please don't use calculus based approach, as I don't know calculus. Also please see my attempt at the solution and correct me.

## Homework Equations

Trigonometric ratios

## The Attempt at a Solution

If the masses are moving downwards with a velocity 'v' then the string is also moving with the same velocity that is 'v' so the velocity with which the mass W should rise should be equal to
'v cos θ' but the correct answer is 'v/cos θ'. Please do tell me why my approach is wrong and what is the correct method

your answer would be right only if V was a force which its obviously isnt
also , if you had the force then the upward force would equal 2F * cos theta . where f is the force at each block and since they look the same then F1=F2
if you had acceleration or time you could apply on this equation F= MA but since you dont , i dont really know , maybe you should law of conservation of momentum but its just a wild guess

I think you are correct and the answer of v/cos theta is wrong.

haruspex
Homework Helper
Gold Member
If the masses are moving downwards with a velocity 'v' then the string is also moving with the same velocity that is 'v'
Quite so, but that means for W to stay connected to each string it must also have a velocity component v towards each pulley. If its speed is u then u cos θ = v.

ehild
Homework Helper
Let be the pieces of string between the pulleys and the middle weight of length L initially. A and B go downward by vΔt in a short time Δt. Then the pieces of the string between the pulley and the middle block get shorter, the new lengths are L'=L-vΔt. The weight raises by Δh.

Apply the Law of Cosine for the blue triangle : L'2=L2+(Δh)2-2LΔh cos(θ), and substitute L'=L-vΔt. Expand, simplify. vΔt and Δh are small, you can ignore their square.Isolate Δh. The speed of the middle block is Δh/Δt.

ehild

#### Attachments

Quite so, but that means for W to stay connected to each string it must also have a velocity component v towards each pulley. If its speed is u then u cos θ = v.

Could you please elaborate I think I'm getting it.

Correct me if I am wrong but I think that the conservation of momentum applies here. The momentum of A and B is must be equal and opposite to the momentum of W.

HallsofIvy
Homework Helper
No, you don't need "conservation of momentum" or any reference to a force. You are told that "the masses are moving downwards with a velocity 'v'" so it is simply a matter of geometry- how does the contraction of the lengths of the sides of the triangle affect the height of the object.

lets see
if the string falls for a distance X in time T
then the Velocity of the strings is V=X/T
and since the string moves X downwards , the strings holding the weight also moves X upwards
thus , raising the weight with X Cos theta .
So Velocity of the weight = X Cos theta / T
and velocity of weight that is falling down = X/T
now the velocity of Weight is equal to X cos theta / t right ?
that is Velocity of Falling body multiplied by Cos theta
so velocity of the body moving upwards is V * cos theta * 2 since T is now half because two bodies are lifting it

ehild
Homework Helper
lets see
if the string falls for a distance X in time T
then the Velocity of the strings is V=X/T
and since the string moves X downwards , the strings holding the weight also moves X upwards
It does not move upward. It gets shorter by X. As the weight W does not change its horizontal position, the angle of the string will change with respect to the vertical. Try to follow the method I suggested, with Law of Cosines.

ehild

Chestermiller
Mentor
In echild's figure in response 5, the three sides of the right triangle to the right are h(t)=Lcosθ, Lsinθ, and L. After a time interval Δt, the lengths of the three sides of the triangle become h(t+Δt), Lsinθ, and L-vΔt. These three sides of the right triangle have to satisfy the Pythagorean theorem, so:

$$h(t+Δt)=\sqrt{(L-vΔt)^2-(L\sinθ)^2}=\sqrt{L^2\cos^2θ-2vΔtL+v^2(Δt)^2}=\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}$$
The average upward velocity of the center weight during the time interval Δt is given by:

$$w_{ave}=\frac{h(t)-h(t+Δt)}{Δt}=\frac{h(t)-\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}}{Δt}$$

If we multiply numerator and denominator of this equation by $h(t)+\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}$, we obtain:

$$w_{ave}=\frac{2v(h(t)/cosθ)+v^2(Δt)}{h(t)+\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}}$$
If we take the limit of this equation as the time interval Δt becomes small, we obtain the instantaneous velocity w:
$$w=\frac{v}{cosθ}$$

ehild
Homework Helper
Nice explanation! I applied the Law of Cosines for the blue triangle in my drawing, but it is the same in principle.

L'2=L2+(Δh)2-2L(Δh)cos(θ)

L'=L-(vΔt), plugging in, expanding, simplifying, ignoring squares of vΔt and Δh, you get the vertical velocity of the weight as u= Δh/Δt.

ehild

Last edited:
haruspex
Homework Helper
Gold Member
In echild's figure in response 5, the three sides of the right triangle to the right are h(t)=Lcosθ, Lsinθ, and L. After a time interval Δt, the lengths of the three sides of the triangle become h(t+Δt), Lsinθ, and L-vΔt. These three sides of the right triangle have to satisfy the Pythagorean theorem, so:

$$h(t+Δt)=\sqrt{(L-vΔt)^2-(L\sinθ)^2}=\sqrt{L^2\cos^2θ-2vΔtL+v^2(Δt)^2}=\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}$$
The average upward velocity of the center weight during the time interval Δt is given by:

$$w_{ave}=\frac{h(t)-h(t+Δt)}{Δt}=\frac{h(t)-\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}}{Δt}$$

If we multiply numerator and denominator of this equation by $h(t)+\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}$, we obtain:

$$w_{ave}=\frac{2v(h(t)/cosθ)+v^2(Δt)}{h(t)+\sqrt{h(t)^2-2v(h(t)/cosθ)Δt+v^2(Δt)^2}}$$
If we take the limit of this equation as the time interval Δt becomes small, we obtain the instantaneous velocity w:
$$w=\frac{v}{cosθ}$$
Chet, ehild, you're making this much too complicated. See my post #4.

haruspex
Homework Helper
Gold Member
lets see
if the string falls for a distance X in time T
then the Velocity of the strings is V=X/T
and since the string moves X downwards , the strings holding the weight also moves X upwards
thus , raising the weight with X Cos theta .
The section of string on the weight's side of the pulley does not move upwards at speed v; it moves towards the pulley at speed v. The end connected to the weight must also be moving towards the pulley at speed v (or the string would change length), as well as moving tangentially, so as to produce a net vertical motion. I.e. the component of its vertical motion that is towards the pulley must be v. Therefore its vertical speed is v sec theta.

Chestermiller
Mentor
Chet, ehild, you're making this much too complicated. See my post #4.
Yes. I like #4 much better. Sorry I missed it the first time through.

Chet

The section of string on the weight's side of the pulley does not move upwards at speed v; it moves towards the pulley at speed v. The end connected to the weight must also be moving towards the pulley at speed v (or the string would change length), as well as moving tangentially, so as to produce a net vertical motion. I.e. the component of its vertical motion that is towards the pulley must be v. Therefore its vertical speed is v sec theta.

i said that it covers Distance X * cos theta * / Time
so V= V * cos theta
will you explain where did you get the sec ?
i thought the component must be Cos since The change in Vertical position is equal to CosTheta the Change in the position of string towards the pully

haruspex