Find the Value of θ in a Pulley Blocks Problem - Homework Solution

[Tanya Sharma]

Homework Statement

Q. In the figure shown, find out the value of θ [ assume string to be tight ]

Homework Equations

The Attempt at a Solution

If the pulley was not moving ,then we could have equated the component of velocities of the two blocks along the string .

But,since the pulley is moving,something extra needs to be done .Not sure what .

I would be grateful if somebody could help me with the problem .

 

[Saitama]

Homework Statement

Q. In the figure shown, find out the value of θ [ assume string to be tight ]

Homework Equations

The Attempt at a Solution

If the pulley was not moving ,then we could have equated the component of velocities of the two blocks along the string .

But,since the pulley is moving,something extra needs to be done .Not sure what .

I would be grateful if somebody could help me with the problem .

I am not sure but how about working in the frame of pulley?

 

[Tanya Sharma]

I am not sure but how about working in the frame of pulley?

Could you elaborate what do you mean by working in the pulley frame ? What benefit does it give in the context of the problem ?

 

[Saitama]

Could you elaborate what do you mean by working in the pulley frame ? What benefit does it give in the context of the problem ?

I meant that if you work in the frame fixed to the pulley, the blocks have an additional component of velocity downwards. You can then equate the velocity along the string. (And that gives one of the answers mentioned.)

 

[Chestermiller]

I analyzed this a little differently. Drop a normal from the pulley to the line joining the tops of blocks A and B. Call the length of this normal h. You then have two right triangles. Express the other sides of these two right triangles trigonometrically in terms of h (using trig functions of 30 degrees and θ). Let a small increment of time Δt pass. Express then new sides of the two triangles in terms of Δt. The sum of the hypotenuses of these new triangles must be the length of the string, which is constrained to be unchanged. So set the sum of the new hypotenuses equal the sum of the original hypotenuses. Expand each of the two hypotenuses as a linear function of Δt. Certain terms will cancel from both sides of the equation. Set the coefficient of Δt equal to zero. Solve for θ, or substitute each of the possible answers into the equation, and see which one satisfies it.

Chet

 

[Tanya Sharma]

I analyzed this a little differently. Drop a normal from the pulley to the line joining the tops of blocks A and B. Call the length of this normal h. You then have two right triangles. Express the other sides of these two right triangles trigonometrically in terms of h (using trig functions of 30 degrees and θ). Let a small increment of time Δt pass. Express then new sides of the two triangles in terms of Δt. The sum of the hypotenuses of these new triangles must be the length of the string, which is constrained to be unchanged. So set the sum of the new hypotenuses equal the sum of the original hypotenuses. Expand each of the two hypotenuses as a linear function of Δt. Certain terms will cancel from both sides of the equation. Set the coefficient of Δt equal to zero. Solve for θ, or substitute each of the possible answers into the equation, and see which one satisfies it.

Chet

Hi Chet...

Sorry for the late response .

Let h be the length of the normal ; x length of the left string and y length of the right string ; a and b be the length of the bases of left and right triangles respectively at t=0.

x=hcosecθ and y=hcosec30°
a=hsecθ and b=hsec30°

Now after time Δt ,

a' = a-3.25Δt
h' = h+Δt
b' = b+√3t

Now x+y = √(a'²+h'²) + √(b'²+h'²)

$$hcosecθ+hcosec30° = \sqrt{(hsecθ-3.25Δt)^2+(h+Δt)^2} + \sqrt{(hsec30°+\sqrt{3}Δt)^2+(h+Δt)^2}$$

Is it correct ?

 

[Chestermiller]

Hi Chet...

Sorry for the late response .

Let h be the length of the normal ; x length of the left string and y length of the right string ; a and b be the length of the bases of left and right triangles respectively at t=0.

x=hcosecθ and y=hcosec30°
a=hsecθ and b=hsec30°

Now after time Δt ,

a' = a-3.25Δt
h' = h+Δt
b' = b+√3t

Now x+y = √(a'²+h'²) + √(b'²+h'²)

$$hcosecθ+hcosec30° = \sqrt{(hsecθ-3.25Δt)^2+(h+Δt)^2} + \sqrt{(hsec30°+\sqrt{3}Δt)^2+(h+Δt)^2}$$

Is it correct ?

Ooops. Those sec's should be cots.

Chet

 

[ehild]

If the pulley was not moving ,then we could have equated the component of velocities of the two blocks along the string .

But,since the pulley is moving,something extra needs to be done .Not sure what .

I would be grateful if somebody could help me with the problem .

You know the motion of everything. The blocks move horizontally, with the given velocities.
The pulley moves vertically upward with 1 m/s.

The string is tight, and it can move on the pulley. The total length is constant.

Choose the origin on the ground, just below the pulley. The positions of the blocks are A (x_a,0) and B(x_B,0). The pulley is at P(0,y). Write up the length of the pieces L_A, L_B. L_A+ L_B=constant. Take the derivative with respect to time. You know the derivative of dx_A/dt=3.25, dX_B/dt=√3, dy/dt=1. y/(-x_A)=tanθ, y/x_B=tan(30°)...


ehild

 

[Tanya Sharma]

Ooops. Those sec's should be cots.

Chet

Yes...poor effort on my part .

If I open the expression under the radical sign ,I get lot of terms none of which seem to cancel.

Now what to do with terms containing Δt and Δt² ?

Expand each of the two hypotenuses as a linear function of Δt. Certain terms will cancel from both sides of the equation. Set the coefficient of Δt equal to zero.

Sorry...I didn't quite understand this .

 

[Chestermiller]

Yes...poor effort on my part .

If I open the expression under the radical sign ,I get lot of terms none of which seem to cancel.

Now what to do with terms containing Δt and Δt² ?



Sorry...I didn't quite understand this .

Discard the (Δt)² terms.

$$\sqrt{(hcotθ-3.25Δt)^2+(h+Δt)^2}≈\sqrt{h^2(cot^2θ+1)+2hΔt(1-3.25cotθ)}≈hcscθ\sqrt{1+\frac{2hΔt(1-3.25cotθ)}{h^2csc^2θ}}$$
$$\sqrt{(hcotθ-3.25Δt)^2+(h+Δt)^2}≈hcscθ+Δt(sinθ-3.25cosθ)$$

Chet

 

[Tanya Sharma]

Thanks ehild ...Very nice and systematic way of looking at the problem.

But I seem to be struggling with sign errors :shy:. I don't get correct answer with the data you have given .Yet it looks fine to me.

Choose the origin on the ground, just below the pulley. The positions of the blocks are A (x_a,0) and B(x_B,0). The pulley is at P(0,y). Write up the length of the pieces L_A, L_B. L_A+ L_B=constant. Take the derivative with respect to time. You know the derivative of dx_A/dt=3.25, dX_B/dt=√3, dy/dt=1. y/(-x_A)=tanθ, y/x_B=tan(30°)...


ehild

If instead of treating (x_a,0) and B(x_B,0) as coordinates of A and B ,I look them as length (distance ) from the origin ,I get correct answer .

Then we would have

dx_A/dt = -3.25, dX_B/dt=√3, dy/dt=1. y/(x_A)=tanθ, y/x_B=tan(30°)

But I don't understand what is the problem with treating x_a and x_B as coordinates instead of lengths(distances).

 

[ehild]

I can not know if you do not show your work...

ehild

 

[TSny]

I like Pranav's approach to this problem. No calculus or expressions for the length of the string needed. In his frame of reference, the pulley has only rotational motion.

 

[ehild]

I like Pranav's approach to this problem. No calculus or expressions for the length of the string needed. In his frame of reference, the pulley has only rotational motion.

Yes, it is very nice...

ehild

 

[utkarshakash]

This problem can be simplified by using Virtual-Work Method.

 

[Tanya Sharma]

Please see this post in conjunction with my earlier post.

$$\sqrt{{x_A}^2+y^2} + \sqrt{{x_B}^2+y^2} = constant$$

$$\frac{2x_A\frac{dx_A}{dt}+2y\frac{dy}{dt}}{2\sqrt{{x_A}^2+y^2}} + \frac{2x_B\frac{dx_B}{dt}+2y\frac{dy}{dt}}{2\sqrt{{x_B}^2+y^2}}$$

$$\frac{\frac{dx_A}{dt}+\frac{y}{x_A}\frac{dy}{dt}}{\sqrt{1+{\frac{y}{x_A}}^2}} + \frac{\frac{dx_B}{dt}+\frac{y}{x_B}\frac{dy}{dt}}{\sqrt{1+{\frac{y}{x_B}}^2}}$$

On putting in values given in post#8 we get

$$\frac{3.25-tanθ}{secθ} + 2 = 0$$ which gives incorrect result .

Whereas if I use values in post#11 I get $\frac{-3.25+tanθ}{secθ} + 2 = 0$ ,which gives correct result .

Could you reflect on this .

 

[utkarshakash]

Please see this post in conjunction with my earlier post.

$$\sqrt{{x_A}^2+y^2} + \sqrt{{x_B}^2+y^2} = constant$$

$$\frac{2x_A\frac{dx_A}{dt}+2y\frac{dy}{dt}}{2\sqrt{{x_A}^2+y^2}} + \frac{2x_B\frac{dx_B}{dt}+2y\frac{dy}{dt}}{2\sqrt{{x_B}^2+y^2}}$$

$$\frac{\frac{dx_A}{dt}+\frac{y}{x_A}\frac{dy}{dt}}{\sqrt{1+{\frac{y}{x_A}}^2}} + \frac{\frac{dx_B}{dt}+\frac{y}{x_B}\frac{dy}{dt}}{\sqrt{1+{\frac{y}{x_B}}^2}}$$

On putting in values given in post#8 we get

$$\frac{3.25-tanθ}{secθ} + 2 = 0$$ which gives incorrect result .

Whereas if I use values in post#11 I get $\frac{-3.25+tanθ}{secθ} + 2 = 0$ ,which gives correct result .

Could you reflect on this .

I think the problem lies in assuming the coordinate of left block as (x_a,0). It should be (-x_a,0) instead, where x_a>0. Now, if you closely look at $d x_a /dt$ which is the rate of change of x_a, you will find that it is -ve. This is because as the left block progresses towards the origin, x_a(which was initially some +ve quantity) approaches zero which clearly shows that x_a is decreasing and a -ve sign must be placed along with 3.25. Thus the second equation is the correct one.

 

[Tanya Sharma]

I think the problem lies in assuming the coordinate of left block as (x_a,0). It should be (-x_a,0) instead, where x_a>0. Now, if you closely look at $d x_a /dt$ which is the rate of change of x_a, you will find that it is -ve. This is because as the left block progresses towards the origin, x_a(which was initially some +ve quantity) approaches zero which clearly shows that x_a is decreasing and a -ve sign must be placed along with 3.25. Thus the second equation is the correct one.

Your reasoning looks quite convincing . Thanks :)

 

[ehild]

You pulled out xA from the square root, which is negative. It should have been |xA|. $\sqrt{(-2)^2}=2$ and not -2.
I pulled out y and got correct result.

ehild

 

[Tanya Sharma]

You pulled out xA from the square root, which is negative. It should have been |xA|.
I pulled out y and got correct result.

ehild

Excellent troubleshooting !

If you had given me one whole day , i wouldn't have found that error :tongue:

 

[ehild]

Excellent troubleshooting !

If you had given me one whole day , i wouldn't have found that error :tongue:

It took me quite a long time, too. Now you will remember to to pull out only absolute value of something from the square root.

ehild