V and the block of mass M starts moving horizontally initially and then moves up. Let at t = 0 the block make an angle of ϕ with the vertical. At t = t 1 the block makes an angle of θ with the vertical.Given it doesn't move upwards till t 1 find the horizontal velocity at t = t 1 .
In the following set up the rope is pulled with a VelocityDetails and Assumptions :
V = 1 5 0 m/s
θ = 3 0 ∘
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Could you tell me what is the virtual work method. I solved using components.
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I don't know its proof but for a closed system ∑ T . V = 0 always holds and ∑ T . 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.
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Is T over here Tension.
Basic Consideration : Velocity along the string is the same everywhere.
Let the required horizontal velocity be V x , then
V x cos 6 0 ∘ = V , or V x = 3 0 0 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 so derivating by t and after simplification we get
R d t d R = x d t d x
Since x R = cos θ 1
By substitution of given value we get the velocity of the block is twice of the rope
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 = 1 5 0 × 2 = 3 0 0
How do you know AB ⊥ BC. Well you are assuming AB = Δ V. So why perpendicular ?
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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
Check out my new wiki Tribonacci sequence
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Simply use virtual work method.
For any close system ∑ T . V = 0
Let the velocity of the block be v 0 (in horizontal direction).
So − T V + T v 0 c o s ( 9 0 − θ ) = 0
So v 0 = 3 0 0 m / s