A 4 kg rock is suspended by a massless string from one end of a 7 m measuring stick. What is the weight of the measuring stick if it is balanced by a support force at the 1 m mark? The acceleration of gravity is 9.81 m/s^2 Answer in units of N.
To solve this problem, we need to consider the torques acting on the measuring stick.
The weight of the rock exerts a torque around the 7 m mark, creating a clockwise rotation. The weight can be calculated using the formula:
Weight = mass * acceleration due to gravity
Weight = 4 kg * 9.81 m/s^2
Weight = 39.24 N
The support force at the 1 m mark creates a counterclockwise torque to balance the weight. The length of the measuring stick from the 1 m mark to the 7 m mark is 6 m. Let's denote the weight of the measuring stick as F, and the distance from the 1 m mark to the center of mass of the measuring stick as d.
Using the formula for torque:
Torque = Force * Distance
For equilibrium, the clockwise torque must be equal to the counterclockwise torque:
Clockwise torque = Counterclockwise torque
Weight * 6 m = F * d
Rearranging the equation to solve for F:
F = (Weight * 6 m) / d
We know that d = 6 m since the center of mass is at the 3.5 m mark (halfway along the measuring stick).
F = (39.24 N * 6 m) / 6 m
F = 39.24 N
Therefore, the weight of the measuring stick is 39.24 N.