Question

A sphere of mass M=3kg and radius R=0 lm is held in place against a wall by
massless string connected to the wall ar an angle 8-37 from the verucal The strmg
connects to the sphere tangentially as shown in the figure. If the magratade of the
normal force (in newton) provided by the wall as 51, then find (g =10m/s)
9
R
R
M
Vauwer​

Answer 1

Explanation:

भाई पहले मैं एक क्वेश्चन पूछ लूंगा उसका आंसर दे दो फिर से आपका मैं इधर दे दूंगा बहुत अच्छा क्वेश्चन है थैंक यू भाई प्लीज मुझे फॉलो करें और मुझे रेट दे मार्ग में ब्रिलिएंट भाई

Answer 2

To find the value of tension in the string, we can consider the forces acting on the sphere. The gravitational force acting on the sphere is given by:

F_gravity = M * g

where M is the mass of the sphere (3 kg) and g is the acceleration due to gravity (10 m/s^2).

Since the sphere is in equilibrium, the sum of forces acting on it must be equal to zero. Hence, the tension in the string must be equal and opposite to the gravitational force acting on the sphere.

Tension = F_gravity = M * g = 3 kg * 10 m/s^2 = 30 N

The tension in the string also provides the normal force acting on the sphere, which is given by:

Normal force = Tension * cos(8-37) = 30 N * cos(8-37) = approximately 27 N

Since the normal force provided by the wall is 51 N, the difference between the normal force and the tension in the string must be equal to the frictional force acting on the sphere:

Friction = Normal force - Tension = 51 N - 30 N = 21 N

So the frictional force acting on the sphere is 21 N.

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