A pendulum consists of an object of mass 1.07 kg that hangs at the end of a massless bar, a distance 1.14 m from the pivot point. Calculate the magnitude of the torque due to gravity about the pivot point if the bar makes an angle of 22° with the vertical.
torque = weightpendulum*cos22*1.14
= 1.07 kg x cos22 x 1.14 kg
= -1.22 kg
= 1.22 kg
Why am I getting this answer incorrect?
Since when is weight measured in kilograms?
I tried Newtons (N) as well
It seems like you made a slight error in your calculation. Let's go through the steps again to find the correct answer.
The torque due to gravity is given by the formula:
torque = weight * lever arm * sin(angle)
In this case, the weight of the pendulum is equal to the gravitational force acting on it, which can be calculated as:
weight = mass * gravity
where:
mass = 1.07 kg (given)
gravity = 9.8 m/s^2 (standard acceleration due to gravity)
Substituting these values, we have:
weight = 1.07 kg * 9.8 m/s^2 = 10.486 kg·m/s^2
Next, we need to calculate the lever arm, which is the perpendicular distance between the pivot point and the line of action of the weight. In this case, the bar has a length of 1.14 m, and it makes an angle of 22° with the vertical. To find the lever arm, we need to calculate the horizontal component of the bar's length:
lever arm = length * cos(angle)
lever arm = 1.14 m * cos(22°)
lever arm = 1.14 m * 0.927
lever arm = 1.055 m
Finally, we can substitute these values back into the torque formula:
torque = 10.486 kg·m/s^2 * 1.055 m * sin(22°)
Using a calculator, we can find that sin(22°) is approximately 0.3746.
torque ≈ 10.486 kg·m/s^2 * 1.055 m * 0.3746
torque ≈ 3.946 N·m
Therefore, the correct magnitude of the torque due to gravity about the pivot point is approximately 3.946 N·m.