The drawing shows a hydraulic chamber in which a spring (spring constant = 1292 N/m) is attached to the input piston, and a rock of mass 24.7 kg rests on the output plunger. The piston and plunger are at the same height, and each has a negligible mass. By how much is the spring compressed from its unstrained position?
To find out how much the spring is compressed from its unstrained position, we need to consider the equilibrium of the forces acting on the system.
Let's assume that the spring is compressed by distance x from its unstrained position. In this case, the input piston exerts a force on the spring given by F = k * x, where k is the spring constant.
According to Newton's third law of motion, the force exerted by the spring on the input piston is equal in magnitude and opposite in direction to the force exerted by the input piston on the spring. Therefore, the force exerted by the input piston is -F = -k * x.
Since there is no vertical motion of the system, the vertical forces must balance each other. The force exerted by the input piston is balanced by the weight of the rock on the output plunger. The weight of the rock is given by Fg = m * g, where m is the mass of the rock and g is the acceleration due to gravity.
Therefore, we have -k * x = m * g.
Rearranging the equation, we can solve for x:
x = -(m * g) / k.
Substituting the given values, m = 24.7 kg, g = 9.8 m/s^2, and k = 1292 N/m:
x = - (24.7 kg * 9.8 m/s^2) / 1292 N/m.
Calculating this expression, we find that the spring is compressed by approximately -0.182 meters or 18.2 centimeters from its unstrained position.
Note that the negative sign indicates that the spring is compressed in the opposite direction of its natural extension.