your gravitational force indeed = m g,
mass in kilograms, g in meters/ second^2, force in Newtons
The torque indeed is the force times the distance to the pivot point,
ASSUMING that the force is perpendicular to the vector from the application point to the pivot point.
hi! can I calculate the torque by simply multiplying the mass, gravity, and distance since torque has a unit of N-m?
Additional context: I taped a coin at one end of the ruler and placed it in the edge of a table until it is on the verge of falling. I took notes of the measurements that I can get. I got the weight of the object, angle of elevation, location of the pivot point, and the distance of the coin from the pivot point. I'm thinking to solve for the torque by multiplying mass and gravity (Newton) and the distance of the coin from the pivot point (m).
3 answers
In your case that distance would be the HORIZONTAL distance from the application point to the pivot point.
For example if your coin were right above the pivot point, the torque would be zero.
For example if your coin were right above the pivot point, the torque would be zero.
m g d cos theta
where theta is the ruler angle above horizontal (your elevation angle) and d is distance of coin from pivot point.
( Do not forget the mass of the ruler at its center)
where theta is the ruler angle above horizontal (your elevation angle) and d is distance of coin from pivot point.
( Do not forget the mass of the ruler at its center)
3 answers