Two-blocks-A-and-B-of-mass-1-kg-and-2-kg-respectively-are-connected-by-a-string-passing-over-a-light-frictionless-pulley-Both-the-blocks-are-resting-on-a-horizontal-floor-and-the-pulley-is-held-such

Question Number 23707 by Tinkutara last updated on 04/Nov/17

Two blocks A and B of mass 1 kg and

2 kg respectively are connected by a

string, passing over a light frictionless

pulley . Both the blocks are resting on a

horizontal floor and the pulley is held

such that string remains just taut . At

time t = 0, a force F = 20 t N, starts

acting on the pulley along vertically

upward direction . Calculate vertical

displacement of the pulley upto the

instant when B loses contact with the

floor .

Commented by Tinkutara last updated on 04/Nov/17

Commented by mrW1 last updated on 05/Nov/17

I will try .

Force in string is T .

T = \frac{F}{2} = \frac{20 t}{2} = 10 t

a_{A} = \frac{T - m_{A} g}{m_{A}} = \frac{10 t - m_{A} g}{m_{A}} = \frac{10 t - 10}{1} = 10 (t - 1)

a_{B} = \frac{T - m_{B} g}{m_{B}} = \frac{10 t - m_{B} g}{m_{B}} = \frac{10 t - 20}{2} = 5 (t - 2)

we see when t ⩾ 1 mass A is off ground

and when t ⩾ 2 mass B is off ground .

s_{P} = displacement of pulley

v_{P} = velocity of pulley

a_{P} = acceleration of pulley

generally a_{P} = \frac{a_{A} + a_{B}}{2}

Phase (1) : 0 ⩽ t ⩽ 1

both blocks are on ground, no motion .

a_{P} = 0, v_{P} = 0, s_{P} = 0 .

Phase (2) : 1 ⩽ t ⩽ 2

block A is off ground with a_{A} = 10 (t - 1) .

block B is still on ground a_{B} = 0 .

a_{P} = \frac{10 (t - 1)}{2} = 5 (t - 1)

\Rightarrow \frac{{dv}_{P}}{dt} = 5 (t - 1)

\int_{0}^{v_{P}} {dv}_{P} = 5 \int_{1}^{t} (t - 1) dt

v_{P} = \frac{5}{2} \times {[{(t - 1)}^{2}]}_{1}^{t} = \frac{5 {(t - 1)}^{2}}{2}

\Rightarrow \frac{{ds}_{P}}{dt} = \frac{5 {(t - 1)}^{2}}{2}

\int_{0}^{s_{P}} {ds}_{P} = \frac{5}{2} \int_{1}^{t} {(t - 1)}^{2} dt

s_{P} = \frac{5}{6} {[{(t - 1)}^{3}]}_{1}^{t} = \frac{5 {(t - 1)}^{3}}{6}

at t = 2

v_{P} (2) = \frac{5}{2} {(2 - 1)}^{2} = \frac{5}{2} m / s

s_{P} (2) = \frac{5}{6} {(2 - 1)}^{3} = \frac{5}{6} m

i . e . as the block B takes off from ground

the pulley has a displacement of \frac{5}{6} m .

Phase (3) : 2 ⩽ t

both blocks are off ground .

a_{P} = \frac{10 (t - 1) + 5 (t - 1)}{2} = \frac{15 (t - 1)}{2}

\frac{{dv}_{P}}{dt} = \frac{15}{2} (t - 1)

\int_{\frac{5}{2}}^{v_{P}} {dv}_{P} = \frac{15}{2} \int_{2}^{t} (t - 1) dt

v_{p} - \frac{5}{2} = \frac{15}{4} [{(t - 1)}^{2} - 1]

\Rightarrow v_{P} = \frac{5}{4} [3 {(t - 1)}^{2} - 1]

\frac{{ds}_{P}}{dt} = \frac{5}{4} [3 {(t - 1)}^{2} - 1]

\int_{\frac{6}{5}}^{s_{P}} {ds}_{P} = \frac{5}{4} \int_{2}^{t} [3 {(t - 1)}^{2} - 1] dt

s_{P} - \frac{5}{6} = \frac{5}{4} [{(t - 1)}^{3} - 1] - \frac{5}{4} (t - 2)

s_{P} = \frac{5}{4} t (t - 1) (t - 2) + \frac{5}{6}

Commented by mrW1 last updated on 05/Nov/17

Commented by Tinkutara last updated on 09/Nov/17

Thank you very much Sir!

Answered by ajfour last updated on 04/Nov/17

l e t a c c . o f p u l l e y b e A,

a c c . o f 1 k g b l o c k i s t h e n 2 A

t e n s i o n i n s t r i n g = \frac{1}{2} (20 t) = 10 t

m o t i o n s t a r t s w h e n T = 10 t ⩾ m g

t h a t i s a t t = 1 s

w h e n 2 k g b l o c k g e t s l i f t e d, 10 t = M g

\Rightarrow t = 2 s

d i s p l a c e m e n t o f p u l l e y = \int_{1}^{2} v d t

v (t) = \int_{1}^{t} A d t

a p p l y i n g F = m a o n 1 k g b l o c k,

T - m g = m a

o r 10 t - g = 2 A \Rightarrow A = \frac{10 t - g}{2}

v e l o c i t y o f p u l l e y v (t) = \int_{1}^{t} (\frac{10 t - g}{2}) d t

v = \frac{{(10 t - g)}^{2}}{40}, S o

d i s p l a c e m e n t s = \int_{1}^{2} \frac{{(10 t - g)}^{2}}{40} d t

s = \frac{{(10 t - g)}^{3}}{1200} ∣_{1}^{2}

i f g = 10 m / s^{2} t h e n s = \frac{5}{6} m .

Commented by Tinkutara last updated on 04/Nov/17

W h y d i s p l a c e m e n t i s n o t \underset{0}{\int^{2}} v d t ? A n d

w h y w e t a k e v (t) o f b l o c k A o n l y ?

Commented by ajfour last updated on 04/Nov/17

u n t i l T e n s i n T = \frac{1}{2} (20 t) < m g

p u l l e y {d o e n}^{'} t m o v e, (t i l l t = 1 s) .

v (t) = \int_{1}^{t} A d t i s v e l o c i t y o f p u l l e y

a t a n y t i m e t . v e l o c i t y o f b l o c k i s

= 2 v = \int_{1}^{t} 2 A d t .

t o f i n d d i s p l a c e m e n t o f p u l l e y

w e n e e d t o i n t e g r a t e v (t) o f

p u l l e y f r o m t = 1 s t o t = 2 s .

Commented by mrW1 last updated on 09/Nov/17

A t t h e m o m e n t t h e f o r c e b e g i n s t o

a p p l y i t s v a l u e i s z e r o . I t n e e d s 1

s e c o n d t i m e t o r e a c h t h e v a l u e 20 N

a n d t h e b l o c k A b e g i n s t o m o v e . T h a t

m e a n s i n t h e f i r s t s e c o n d i n w h i c h

t h e f o r c e a p p l i e s t h e r e i s n o m o v e m e n t

o f t h e p u l l e y : v = 0 f o r t = 0 t o 1 .

s (t) = \int_{0}^{t} v d t = \int_{0}^{1} v d t + \int_{1}^{t} v d t = 0 + \int_{1}^{t} v d t = \int_{1}^{t} v d t

Commented by Tinkutara last updated on 09/Nov/17

Thank you very much Sir!

The pulley moves only when either of

the two blocks lifts up ? Why not from

the moment when force is applied on it ?

Commented by Tinkutara last updated on 09/Nov/17

B e f o r e 1 s e c o n d, s a y 0.5 s e c o n d, f o r c e

a p p l i e d i s 10 N o n t h e p u l l e y a n d i t i s

c o n t i n u o u s l y i n c r e a s i n g . S o w h y

p u l l e y d o e s n o t m o v e w i t h t h i s f o r c e ?

Commented by mrW1 last updated on 09/Nov/17

I n t h e f i r s t s e c o n d t h e f o r c e F {d o e n}^{'} t

m a k e m o v e m e n t b u t o n l y r e d u c e t h e

t h e c o n t a c t f o r c e N b e t w e e n t h e b l o c k A

a n d t h e f l o o r :

N = m_{A} g - F

T h e b l o c k A w i l l b e l i f t e d w h e n N = 0 .

Commented by Tinkutara last updated on 09/Nov/17

OK . But when block 1 is lifted at 1 s

block 2 is not lifted . So why the pulley

still moves without lifting of block 2 ?

Commented by mrW1 last updated on 09/Nov/17

S i n c e F m u s t b e 40 N t o l i f t t h e b l o c k

B . S o i n t h e t i m e t = 1 t o 2 o n l y b l o c k

A m o v e s . W h e n b l o c k A m o v e s, t h e

p u l l e y m o v e s a l s o .

Commented by Tinkutara last updated on 09/Nov/17

Thank you very much Sir!

All doubts cleared .