Question Video: Finding the Tension in a String Attaching a Sphere to a Wall- and the Normal Reaction with the Wall | Nagwa Question Video: Finding the Tension in a String Attaching a Sphere to a Wall- and the Normal Reaction with the Wall | Nagwa

# Question Video: Finding the Tension in a String Attaching a Sphere to a Wall- and the Normal Reaction with the Wall Mathematics • Second Year of Secondary School

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A light string is tied from one end to a point on the surface of a homogeneous sphere, and the other end is attached to a point on a vertical smooth wall. The sphere is resting against the wall and weighs 33 N, and the string inclines to the vertical by 30Β°. Find the tension π in the string and the reaction π of the wall.

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### Video Transcript

A light string is tied from one end to a point on the surface of a homogeneous sphere, and the other end is attached to a point on a vertical smooth wall. The sphere is resting against the wall and weighs 33 newtons, and the string inclines to the vertical by 30 degrees. Find the tension π in the string and the reaction π of the wall.

In this question, we have a sphere resting against a smooth vertical wall. And the weight of the sphere is 33 newtons. This means that this force acts vertically downwards. The reaction force π acts perpendicular to the wall. Therefore, it acts horizontally to the right. We have a third force π, which is the tension in the string. The reaction force and the weight of the sphere act at right angles to one another. And we are also told that the string inclines to the vertical by 30 degrees. Since angles in a triangle sum to 180 degrees, the third angle in our triangle is 60 degrees.

Since there are three forces acting at a point, we can solve this problem using Lamiβs theorem. This states that when three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces. This can be written as follows. 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 three forces acting at a point. πΌ is the angle between forces π΅ and πΆ; π½, the angle between forces π΄ and πΆ; and πΎ, the angle between the forces π΄ and π΅.

In this question, we already know that the angle between the 33-newton force and π is 90 degrees. By adding 90 degrees to 60 degrees, the angle between the 33-newton force and the tension π is 150 degrees. And finally, since angles at a point sum to 360 degrees, the angle between the tension and reaction forces is 120 degrees.

We now have three forces and their corresponding angles that we can substitute into Lamiβs theorem. This gives us π over sin of 90 degrees is equal to π over sin of 150 degrees which is equal to 33 over sin of 120 degrees. The sin of 90 degrees is equal to one, the sin of 150 degrees is one-half, and the sin of 120 degrees is root three over two. Dividing by a half is the same as multiplying by two. Therefore, π is equal to two π.

The third part of our equation is 33 divided by root three over two. Multiplying the numerator and denominator by two, this is equivalent to 66 over root three. We can then rationalize the denominator by multiplying the numerator and denominator by root three. This gives us 66 root three over three, which in turn is equal to 22 root three. Simplifying all three terms in Lamiβs theorem gives us π is equal to two π which is equal to 22 root three.

The tension in the string is therefore equal to 22 root three newtons. As two π is equal to 22 root three, π is equal to 11 root three. The reaction force of the wall is 11 root three newtons. We have now calculated the values of the missing forces in the question.

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