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?
The answer will depend upon the ratio of the piston areas. The large piston will exert a larger force. Were piston areas or diameters indicated in the drawing you did not show?
dragonballs
To determine how much the spring is compressed, we need to consider the forces acting on it.
First, let's understand the forces acting on the spring system. We have two forces: the force exerted by the spring (F_spring) and the gravitational force acting on the rock (F_gravity).
The force exerted by the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The equation for Hooke's Law is:
F_spring = -k * x
Where:
F_spring is the force exerted by the spring,
k is the spring constant (1292 N/m in this case),
x is the displacement of the spring from its unstrained position (what we want to find).
The gravitational force acting on the rock is given by:
F_gravity = m * g
Where:
F_gravity is the gravitational force,
m is the mass of the rock (24.7 kg in this case),
g is the acceleration due to gravity (9.8 m/s^2).
The force exerted by the spring (F_spring) and the gravitational force (F_gravity) are equal when the system is in equilibrium. Since the piston and plunger are at the same height, the displacement of the spring (x) is equal to the compression of the spring (c) due to the weight of the rock.
Equating F_spring and F_gravity, we have:
-k * x = m * g
Rearranging the equation to solve for x, we get:
x = -m * g / k
Plugging in the given values:
x = - (24.7 kg) * (9.8 m/s^2) / (1292 N/m)
Solving this equation will give us the answer: the amount of spring compression from its unstrained position.