# Problem related to pulley system and velocity

• hasnainzeenwa
In summary, echild's diagram shows that if the masses A and B are moving downwards with a velocity v, then the mass W will also move downwards with a velocity v/cos θ.

## 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 isn't
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 don't , i don't 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.

hasnainzeenwa said:
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.

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

• pulleys.JPG
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haruspex said:
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.

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

B4ssHunter said:
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

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:
$$w=\frac{v}{cosθ}$$

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

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Chestermiller said:
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.

B4ssHunter said:
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.

haruspex said:
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

haruspex said:
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

The length of the string can not change. If the vertical piece gets longer by X, the other one, connected to the weight W is shortened by X. So the end of the string has to move upward, its direction will change. See attachment.

ehild

#### Attachments

• shortstr.JPG
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B4ssHunter said:
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
In time dt it must be v dt closer to (each) pulley. Its net movement must be vertical. If it moves upwards distance dx and as a result is v dt closer to a pulley then dx cos theta = v dt. Thus dx/dt = v / cos theta = v sec theta.