A uniform rod of mass M = 326 kg is in the shape of a semicircle of radius R = 6 m. Calculate the magnitude of the force on a point of mass m = 74 kg placed at the center of the semicircle, as shown in the figure.
The angular mass density is M/pi, the mass between theta and theta + dtheta is thus M/pi dtheta. If the two ends of the semiciscle are at -pi/2 and pi/2, then the foce will be in the theta = 0 direction. The component in that direction from a mass element at theta is obtained by multiplying the force from that direction by cos(theta).
The magnitude of the force is thus given by:
M/pi m G/R^2 Integral from -pi/2 to pi/2 of cos(theta) dtheta =
2/pi M m G/R^2
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To calculate the magnitude of the force on a point mass placed at the center of the semicircle, we need to consider the gravitational force.
The gravitational force acting on an object is given by the equation F = m * g, where F is the force, m is the mass of the object, and g is the acceleration due to gravity.
In this case, the mass of the point mass is m = 74 kg. The acceleration due to gravity, g, is approximately 9.8 m/s^2.
So, the force acting on the point mass is F = m * g = 74 kg * 9.8 m/s^2 = 725.2 N.
Therefore, the magnitude of the force on the point mass placed at the center of the semicircle is 725.2 Newtons.