Question

a smooth sphere of weight W is supported by a string fastened to a point A on the smooth vertical wall, the other end is in contact with point B on the wall . if length of string AC is equal to radius of the sphere of the wall.​

Answer 1

Answer:

Weight (W) of the sphere acts in the downward direction, rope exerts a tension force (T) along its length and the wall exerts a normal force (N) on the sphere as shown in the figure.

Explanation:

Pls Mark Me As Brainliest.

Answer 2

Answer: Tension in the string is 1.15W and normal reaction is 0.577W.

Given: Sphere of weight W.

To Find: Tension in the string and normal reaction of the wall.

Step-by-step explanation:

Step 1: Normal response is the force that a surface applies to an object that comes into touch with it in order to keep the object from piercing the surface. In the absence of frictional forces, the surface only exerts only one force on the item, which is perpendicular to the surface.

It will form a right angle triangle with base of length r and hypotenuses 2r.

Thus,

$Cos\ \theta = \frac{r}{2r} \\\\cos\ \theta = 0.5\\\\\theta = 60$

Step 2: There is no net force exerted on an object and it is considered to be in equilibrium if the amount and direction of the forces acting on it are precisely balanced.

So, using equilibrium of all the forces,

$T sin\ 60 = W\\\\T cos\ 60 = N\\\\T=\frac{W}{sin\ 60}=1.15W \\\\N=T cos\ 60=1.15W\times\frac{1}{2} N=0.577W$

Hence, Tension in the string is 1.15W and normal reaction is 0.577W.

For more questions please follow the link given below.

https://brainly.in/question/1151639

https://brainly.in/question/40839426  

#SPJ2