Four acrobats of mass 83.5 kg, 68.0 kg, 62.0 kg, and 55.0 kg form a human tower, with each acrobat standing on the shoulders of another acrobat. The 83.5-kg acrobat is at the bottom of the tower.
(a) What is the normal force acting on the 83.5-kg acrobat?
Fn = m*g = (83.5+68+62+55) * 9.8
To find the normal force acting on the 83.5 kg acrobat, we need to consider the forces acting on the acrobat in the vertical direction.
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the surface supporting the 83.5 kg acrobat is the shoulders of the acrobat below.
We know that the weight of an object is equal to its mass multiplied by the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 meters per second squared (m/s^2).
The weight of the 83.5 kg acrobat is given by:
Weight = mass * acceleration due to gravity
Weight = 83.5 kg * 9.8 m/s^2
Weight = 819.3 Newtons (N)
Since the 83.5 kg acrobat is at the bottom of the tower, there is no other force pushing it down apart from its weight. Therefore, the normal force acting on the 83.5 kg acrobat is equal to its weight, which is 819.3 N.