A sphere of mass 30.0 kg and radius 0.150 m is attached to a point on a smooth vertical wall by a string of length 0.150 m which runs from a point on the surface of the sphere to the point on the wall. Find the tension in the string and the horizontal force applied to the sphere by the wall
To find the tension in the string and the horizontal force applied to the sphere by the wall, we need to analyze the forces acting on the sphere.
Let's start by considering the forces acting on the sphere when it is in equilibrium. There are two forces acting on the sphere: the weight of the sphere and the tension in the string.
1. Weight of the sphere (W):
The weight of an object is given by the formula W = mg, where m is the mass of the object and g is the acceleration due to gravity. In this case, the mass of the sphere is given as 30.0 kg, and we'll assume the acceleration due to gravity is 9.8 m/s².
Thus, the weight of the sphere is:
W = (30.0 kg) * (9.8 m/s²) = 294 N (Newtons)
2. Tension in the string (T):
The tension in the string is the force exerted by the string to keep the sphere in equilibrium. In this case, the string is attached at a point on the surface of the sphere and to a point on the wall.
Since the string is connected to a point on the wall, it can provide a horizontal force. Let's call this horizontal force F.
As the sphere is in equilibrium, the vertical component of the tension in the string and the weight of the sphere must be balanced. This means T (vertical) = W.
Since the string is attached to the sphere and forms a right-angled triangle with the vertical wall, we can use trigonometry to find the vertical component of the tension.
The vertical component of the tension is given by the formula T (vertical) = T * sin θ, where θ is the angle formed between the string and the vertical wall.
In this case, the length of the string is given as 0.150 m, and the radius of the sphere is also given as 0.150 m. So the angle θ is equal to the angle formed between the string and the vertical wall, which is 90 degrees.
Thus, sin θ = sin 90° = 1.
Therefore, T (vertical) = T * 1 = T.
Since T (vertical) = W, we can equate the two:
T = W = 294 N.
So the tension in the string is 294 N.
To find the horizontal force applied to the sphere by the wall, we need to find the horizontal component of the tension.
The horizontal component of the tension is given by the formula T (horizontal) = T * cos θ.
In this case, cos θ = cos 90° = 0.
Therefore, T (horizontal) = T * 0 = 0 N.
Hence, the horizontal force applied to the sphere by the wall is 0 N.