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in Physics by (15 points)
A light string ABCDE whose extremity A is fixed, has weights W1 and W2 attached to it at B
and C. It passes round a small smooth peg at D carrying a weight of 300 N at the free end E
as shown in figure 5.0.
 Figure 5.0
 If the system is in equilibrium, determine;
i. The tensions in the portion AB, BC and CD.
ii. The magnitudes of W1 and W2.

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1 Answer

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by (48.9k points)

First string ABCD is split in to two parts, and consider the joints B and C separately Let

T1 = Tension in String AB

T2 = Tension in String BC

T3 = Tension in String CD

T4 = Tension in String DE 

T4 = T3 = 40 N

Since at joint B and C three forces are acting on both points. But at B all three forces are unknown and at point C only two forces are unknown 

SO Apply lamis theorem first at joint C,

T2 /sin 150° = W2/sin 120° = 40/sin 90°

= 20N

T2 = {sin 150° × 40}/sin 90°

W2 = {sin 120° × 40}/sin 90°

= 34.64N

Now for point B, We know the value of T2 

So, Again Apply lamis theorem at joint 

B, T1/sin 90° = W1/sin 150° = T2/sin 120°

T1 = {sin 90° × 20}/sin 120° 

= 23.1 N

W1 = {sin 150° × 20}/sin 120°

= 11.55N

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