Magnitude of Torgue: Due To Force of Gravity?

Question

A simple pendulum consists of 3.1 kg point mass hanging at the end of 2.0 m long light string that is connected to a pivot point.
Calculate the magnitude of torque, due to the force of gravity, around this pivot point when the string makes a 2.8degree angle with the vertical. Accleleration of gravity is 9.81 m/s^2. Answer in units of N m.

What would the calculation be for a 19.9degree angle? Answer in units of N m.

Answer 1

Torque is the cross product of force and distance.
T = F x d

This is often written using a trig function
T = F d sin H where H is the angle between F and d.

In this case, since we can neglect the "light string" the only force is mg = (3.1 kg)(9.8 m/s^2). The distance d is 2.0 m. The torque is then computed as

(3.1 kg)(9.8 m/s^2) (2.0 m) (sin H) where H is the angle made with the vertical.

Plug in your two angles to get your two answers.

Answer 2

rigidity is measured in newtons. Acceleration is meters in keeping with 2d squared. that's incomprehensible to evaluate the numbers for them. A a million kg merchandise resting on a floor could have a rigidity of 9.8 newtons. it particularly is going to fall with an acceleration of 9.8 m/s^2. Then a 1000 kg merchandise resting on a floor could have a rigidity of 9800 newtons. it additionally will fall with an acceleration of 9.8 m/s^2.

Answer 3

T= r x F
F=mgsin(Theta)
T=r mgsin(2.8)
T=2.0 x 3.1x 9.81 x sin(2.8)=
T=2.97 Nm

and for 19.9 degrees
T=2.0 x 3.1x 9.81 x sin(19.9)=
T=20.7 Nm