Question Video: Analysis of a Weight Suspended by Two Strings Inclined to the Vertical | Nagwa Question Video: Analysis of a Weight Suspended by Two Strings Inclined to the Vertical | Nagwa

# Question Video: Analysis of a Weight Suspended by Two Strings Inclined to the Vertical Mathematics • Second Year of Secondary School

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A weight of 90 g-wt is suspended by two inextensible strings. The first is inclined at an angle π to the vertical, and the second is at 30Β° to the vertical. If the magnitude of the tension in the first string is 45 g-wt, find π and the magnitude of the tension π in the second string.

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### Video Transcript

A weight of 90 gram-weight is suspended by two inextensible strings. The first is inclined at an angle π the vertical, and the second is at 30 degrees to the vertical. If the magnitude of the tension in the first string is 45 gram-weight, find π and the magnitude of the tension π in the second string.

We will begin here by sketching a diagram. We are told that a 90 gram-weight is suspended by two inextensible strings as shown. Theyβre inclined at angles π and 30 degrees to the vertical. The magnitude of the tension in the first string is 45 gram-weight. And we need to calculate the magnitude of the tension π in the second string. As angles on a straight line sum to 180 degrees, the angle between the tension π and the 90 gram-weight is 150 degrees. In the same way, the angle between the 45 gram-weight tension and the 90 gram-weight is 180 minus π degrees.

As we have three coplanar, concurrent, and noncollinear forces acting to keep the object in static equilibrium, we can use Lamiβs theorem. This states that π΄ over sin πΌ is equal to π΅ over sin π½, which is equal to πΆ over sin πΎ, where π΄, π΅, and πΆ are the magnitude of the forces and πΌ, π½, and πΎ are the angles opposite them. In this question, the three forces will be π, 90, and 45. The angles will be 180 minus π, π plus 30, and 150. Substituting in these values, we have π over sin of 180 minus π is equal to 90 over sin of π plus 30 which is equal to 45 over sin of 150.

Letβs consider the second and third parts to this equation. sin of 150 degrees is equal to one-half. So, our equation becomes 90 over sin of π plus 30 is equal to 45 over a half. 45 divided by a half is equal to 90. And we can then multiply both sides by the sin of π plus 30. Dividing both sides of this new equation by 90 gives us one is equal to the sin of π plus 30. We can then take the inverse sin of both sides of this equation. The inverse sin of one is equal to 90 degrees. So, 90 is equal to π plus 30. Subtracting 30 from both sides of this equation gives us π is equal to 60. The first part of our answer is π is equal to 60 degrees.

We can now substitute this value back into our equation to help calculate π. Using the first two parts of our equation, we have π over sin of 120 degrees is equal to 90 over sin of 90 degrees. sin of 90 degrees is equal to one, and we can then multiply both sides by the sin of 120 degrees. π is equal to 90 multiplied by the sin of 120 degrees. The sin of 120 degrees is equal to root three over two. Multiplying this by 90, we get π is equal to 45 root three. The tension in the second string is equal to 45 root three gram-weight. The two answers to this question are π equals 60 degrees and π is equal to 45 root three gram-weight.

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