150 Follower Problem!

In the following set up the rope is pulled with a Velocity V V and the block of mass M M starts moving horizontally initially and then moves up. Let at t = 0 t = 0 the block make an angle of ϕ \phi with the vertical. At t = t 1 t = t_1 the block makes an angle of θ \theta with the vertical.Given it doesn't move upwards till t 1 t_1 find the horizontal velocity at t = t 1 t = t_1 .

Details and Assumptions :

  • V = 150 m/s V = 150 \text{ m/s}

  • θ = 3 0 \theta = 30^{\circ}


The answer is 300.

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4 solutions

Satvik Pandey
Mar 30, 2015

Simply use virtual work method.

For any close system T . V = 0 \sum { \vec { T } .\vec { V } } =0

Let the velocity of the block be v 0 v_{0} (in horizontal direction).

So T V + T v 0 c o s ( 90 θ ) = 0 -TV+Tv_{0}cos(90-\theta)=0

So v 0 = 300 m / s v_{0}=300 m/s

Could you tell me what is the virtual work method. I solved using components.

Rajdeep Dhingra - 6 years, 2 months ago

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I don't know its proof but for a closed system T . V = 0 \sum { \vec { T } .\vec { V } } =0 always holds and T . A = 0 \sum { \vec { T } .\vec { A } } =0 also holds but there are some situations in which the latter one does not holds. Watch this video to know more about this.

satvik pandey - 6 years, 2 months ago

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Is T over here Tension.

Rajdeep Dhingra - 6 years, 2 months ago

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@Rajdeep Dhingra Yes, it is Tension.

satvik pandey - 6 years, 2 months ago
Swapnil Das
Dec 30, 2016

Basic Consideration : Velocity along the string is the same everywhere.

Let the required horizontal velocity be V x {V}_{x} , then

V x cos 60 = V { V }_{ x }\cos { { 60 }^{ \circ } } =V , or V x = 300 m/s { V }_{ x }=300\quad \text{m/s} .

Considering H as the height of pulley, R the lenght of the rope and x the distance of the block to the vertical of the pulley, being R=R(t) , x=x(t) and H equal a constant. Now we can write by Phytagoras

R 2 = H 2 + x 2 R^{2}=H^{2}+x^{2} so derivating by t and after simplification we get

R d R d t = x d x d t R\frac{dR}{dt} =x\frac{dx}{dt}

Since R x = 1 cos θ \frac{R}{x} = \frac{1}{\cos \theta}

By substitution of given value we get the velocity of the block is twice of the rope

Archit Boobna
Mar 31, 2015

T h i s f i g u r e s h o w s t h e p o s i t i o n o f t h e b l o c k a f t e r a n e g l e g i b l e t i m e Δ . S i n c e t h e v e l o c i t y o f t h e r o p e i s V , i t s l e n g t h o n t h e l e f t s i d e o f t h e p u l l e y c h a n g e s b y Δ . V i n Δ t i m e . S o A B = Δ . V ( 1 ) A C i s t h e d i s p l a c e m e n t c o v e r e d i n Δ t i m e b y t h e b l o c k . S o , A C = Δ . U ( W h e r e U i s t h e v e l o c i t y o f t h e b l o c k ) ( 2 ) B y s i m p l e t r i g o w e c a n s e e t h a t A C = A B csc θ ( 3 ) U s i n g ( 1 ) , ( 2 ) a n d ( 3 ) , w e g e t Δ . U = Δ . V csc θ S o U = V csc θ S o U = 150 × 2 = 300 This\quad figure\quad shows\quad the\quad position\quad of\quad the\quad block\quad after\quad a\\ neglegible\quad time\quad \Delta .\\ \\ Since\quad the\quad velocity\quad of\quad the\quad rope\quad is\quad V,\quad its\quad length\quad\\ on\quad the\quad left\quad side\quad of\quad the\quad pulley\quad changes\quad by\quad \Delta .V\quad in\quad \Delta \quad time.\\ So\quad AB=\Delta .V\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left( 1 \right) \\ \\ AC\quad is\quad the\quad displacement\quad covered\quad in\quad \Delta \quad time\quad by\quad the\quad block.\\ \\ So,\quad AC=\Delta .U\quad (Where\quad U\quad is\quad the\quad velocity\quad of\quad the\quad block)\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( 2 \right) \\ By\quad simple\quad trigo\quad we\quad can\quad see\quad that\quad AC=AB\csc { \theta } \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left( 3 \right) \\ \\ Using\quad \left( 1 \right) ,\left( 2 \right) \quad and\quad \left( 3 \right) ,\quad we\quad get\\ \Delta .U=\Delta .V\csc { \theta } \\ So\quad U=V\csc { \theta } \\ So\quad U=150\times 2=300

How do you know AB \perp BC. Well you are assuming AB = Δ \Delta V. So why perpendicular ?

Rajdeep Dhingra - 6 years, 2 months ago

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Let D be the point of intersection of rope and pulley

We assume that the triangle BCD is isosceles. And angle BDC is 0 because the time is negligible, so ABC=BCD=90

Archit Boobna - 6 years, 2 months ago

Check out my new wiki Tribonacci sequence

Archit Boobna - 6 years, 2 months ago

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