A meter stick has mass 39 g, and has a pivot placed at the center at the 50 cm mark. A mass of m1 is place at the 5 cm mark. Another mass of 10 g is placed at the 70 cm mark. What mass m in grams has to be placed at the 80 cm mark to create translational and rotational equilibrium? Hint: To solve this problem you have to use the equation from Newton's second law set equal to zero and the rotational equilibrium equation.
Not sure how to do this one. aren't the newtons 2nd law and and rotational equilibrium equations the same?
47gram
How did you come to that answer please?
No, Newton's second law and the equation for rotational equilibrium are not the same. Let me explain both concepts and how they can be used to solve this problem.
Newton's second law deals with the net force acting on an object and its resulting acceleration. The formula for Newton's second law is given as:
F = ma
Where F is the net force applied on the object, m is its mass, and a is the resulting acceleration. In this problem, we are interested in finding the mass that needs to be placed at the 80 cm mark to create equilibrium, so we need to set up the equation for translational equilibrium using Newton's second law.
On the other hand, the equation for rotational equilibrium deals with the net torque acting on an object and its resulting angular acceleration. The formula for rotational equilibrium is:
τ = Iα
Where τ is the net torque, I is the moment of inertia of the object, and α is the resulting angular acceleration. To solve this problem, we will use the equation for rotational equilibrium along with the equation for translational equilibrium.
Let's break down the steps to solve the problem:
1. Set up the equation for translational equilibrium using Newton's second law:
Sum of forces = 0
In this case, the only forces acting on the meter stick are the weights of the masses. So we have:
m1g + m2g + mg = 0
Here, g is the acceleration due to gravity, m1 is the mass at the 5 cm mark, m2 is the mass at the 70 cm mark, and m is the mass we need to find at the 80 cm mark.
2. Set up the equation for rotational equilibrium:
Sum of torques = 0
For rotational equilibrium, the sum of the torques must be zero. In this case, we need to calculate the torque due to each mass about the pivot point at the center.
The torque due to each mass is given by the formula:
τ = rF
Where r is the distance of the mass from the pivot point and F is the force due to the weight of the mass.
So we have:
(r1)(m1g) + (r2)(m2g) + (r3)(mg) = 0
Here, r1 is the distance from the pivot to the 5 cm mark, r2 is the distance from the pivot to the 70 cm mark, and r3 is the distance from the pivot to the 80 cm mark.
3. Solve the two equations simultaneously to find the mass m:
Substitute the values of the distances and masses into the equations for translational and rotational equilibrium. Solve for m.
By solving these two equations simultaneously, you can find the mass m that needs to be placed at the 80 cm mark to create translational and rotational equilibrium.