six forces lying in a plane and forming angles of 60°relative to one another are applied the centre of a homogenous sphere with a mass m=6kg. these forces are radially outward and consecutively 1N,2N,3N,4N,5N,6N. THE acceleration of the sphere is?answer is 1m/s² can u write the procedure
Answers
Try to resolve force in opposite direction
then resultant will have three force each of 3N at 0o, 60o, 120o
then resulatnt of it will have two force of 3N and 3\sqrt{3} at angle 90
then resultant will be \sqrt{(3\sqrt{3})^{2} + 3^{2}} = 6N
Then acceleration = 6/6 = 1m/s2
Hope it clears.................................
The acceleration of the sphere is 1 m/s².
Given: six forces lying in a plane and forming angles of 60°relative to one another. The magnitude of the forces is 1 N, 2 N, 3 N, 4 N, 5 N, and 6 N.
The mass of the sphere is 60 kg.
To Find: The acceleration of the sphere
Solution:
- The sphere is divided into 6 parts and along each part, the magnitude of the force is consecutively 1 N, 2 N, 3 N, 4 N, 5 N, and 6 N.
- If we visualize the sphere, we can observe that we have 3 of these forces, acting diametrically opposite to each other, which means they lie in the same plane.
- Now we need to find the resultant of the forces.
OA = 6 - 3 = 3 N
OB = 5 - 2 = 3 N
OC = 4 - 1 = 3 N [ where O is the centre of the circle ]
- After calculating, we can observe that we have 3 forces, each of magnitude 3 N and inclined at each other at an angle of 60°.
- For finding the acceleration, we shall use the formula,
∑ F = m × a
Coming to the numerical, we need to find the magnitude of the component of the forces along the x-direction and y-direction.
So, Fx = 3 + 3 cos 60° + 3 cos 120°
= 3 + 3/2 - 3/2
= 3 N
Now, Fy = 3 cos 30° + 3 cos 30°
= 2 × 3 × √3/2
= 3√3 N
Now, F = √ ( Fx² + Fy² )
= √ ( 3² + ( 3√3 )² )
= √36 N
= 6 N
Again, we know that F = m × a and m = 6 kg.
So, a = F / m = ( 6 / 6 ) m/s²
= 1 m/s²
Hence, the acceleration of the sphere is 1 m/s².
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