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Question: Two smooth beads A and B, free to move on a vertical smooth circular wire, are connected by a string...

Two smooth beads A and B, free to move on a vertical smooth circular wire, are connected by a string. Weights W1, W2 and W are suspended from A, B and a point C of the string respectively. In equilibrium, A and B are in a horizontal line. If ∠BAC=α\angle B A C = \alpha and ∠ABC=β\angle A B C = \beta , then the ratio tan⁡α:tan⁡β\tan \alpha : \tan \beta is

A

tan⁡αtan⁡β=W−W1+W2 W+W1−W2\frac { \tan \alpha } { \tan \beta } = \frac { \mathrm { W } - \mathrm { W } _ { 1 } + \mathrm { W } _ { 2 } } { \mathrm {~W} + \mathrm { W } _ { 1 } - \mathrm { W } _ { 2 } }

B

tan⁡αtan⁡β=W+W1−W2W−W1+W2\frac { \tan \alpha } { \tan \beta } = \frac { W + W _ { 1 } - W _ { 2 } } { W - W _ { 1 } + W _ { 2 } }

C

tan⁡αtan⁡β=W+W1+W2W+W1−W2\frac { \tan \alpha } { \tan \beta } = \frac { W + W _ { 1 } + W _ { 2 } } { W + W _ { 1 } - W _ { 2 } }

D

None of these

Answer

tan⁡αtan⁡β=W−W1+W2 W+W1−W2\frac { \tan \alpha } { \tan \beta } = \frac { \mathrm { W } - \mathrm { W } _ { 1 } + \mathrm { W } _ { 2 } } { \mathrm {~W} + \mathrm { W } _ { 1 } - \mathrm { W } _ { 2 } }

Explanation

Solution

Resolving forces horizontally and vertically at the points A, B and C respectively, we get

Tcos⁡α=R1sin⁡γT \cos \alpha = R _ { 1 } \sin \gamma .....(i)

T1sin⁡α+W1=R1cos⁡γT _ { 1 } \sin \alpha + W _ { 1 } = R _ { 1 } \cos \gamma .....(ii)

T1cos⁡β=R2sin⁡γT _ { 1 } \cos \beta = R _ { 2 } \sin \gamma .....(iii)

T2sin⁡β+W2=R2cos⁡γT _ { 2 } \sin \beta + W _ { 2 } = R _ { 2 } \cos \gamma .....(iv)

T1cos⁡α=T2cos⁡βT _ { 1 } \cos \alpha = T _ { 2 } \cos \beta .....(v)

and T1sin⁡α+T2sin⁡β=WT _ { 1 } \sin \alpha + T _ { 2 } \sin \beta = W .....(vi)

Using (v), from (i)and (ii), we get, R1=R2R _ { 1 } = R _ { 2 }

∴ From (ii) and (vi), we have

T1sin⁡α+W1=T2sin⁡β+W2T _ { 1 } \sin \alpha + W _ { 1 } = T _ { 2 } \sin \beta + W _ { 2 }

or T1sin⁡α−T2sin⁡β=W2−W1T _ { 1 } \sin \alpha - T _ { 2 } \sin \beta = W _ { 2 } - W _ { 1 } .....(vii)

Adding and subtracting (vi) and (vii), we get

2T1sin⁡α=W+W2−W12 T _ { 1 } \sin \alpha = W + W _ { 2 } - W _ { 1 } ......(viii)

2T2sin⁡β=W−W2+W12 T _ { 2 } \sin \beta = W - W _ { 2 } + W _ { 1 } ......(ix)

Dividing (viii) by (ix), we get

T1T2⋅sin⁡αsin⁡β=W−W1+W2W+W1−W2\frac { T _ { 1 } } { T _ { 2 } } \cdot \frac { \sin \alpha } { \sin \beta } = \frac { W - W _ { 1 } + W _ { 2 } } { W + W _ { 1 } - W _ { 2 } } or cos⁡βcos⁡α⋅sin⁡αsin⁡β=W−W1+W2W+W1−W2\frac { \cos \beta } { \cos \alpha } \cdot \frac { \sin \alpha } { \sin \beta } = \frac { W - W _ { 1 } + W _ { 2 } } { W + W _ { 1 } - W _ { 2 } } (from (v)) or tan⁡αtan⁡β=W−W1+W2W+W1−W2\frac { \tan \alpha } { \tan \beta } = \frac { W - W _ { 1 } + W _ { 2 } } { W + W _ { 1 } - W _ { 2 } }