A force of 60 N is exerted on one end of a 1.0-m-long lever. The other end of the lever is attached to a rotating rod that is perpendicular to the lever. By pushing down on the end of the lever, you can rotate the rod. If the force on the lever is exerted at an angle of 30degrees, what torque is exerted on the lever?
.. an angle of 30 degrees relative to what direction?
To find the torque exerted on the lever, we need to calculate the moment of force or torque using the formula:
Torque = Force × Distance × sin(θ)
where:
Force = 60 N (given)
Distance = 1.0 m (given)
θ = 30° (given)
Now, let's substitute the given values into the formula to find the torque:
Torque = 60 N × 1.0 m × sin(30°)
The sin(30°) can be calculated as 0.5, so we have:
Torque = 60 N × 1.0 m × 0.5
= 30 Nm
Therefore, the torque exerted on the lever is 30 Nm.
To calculate the torque exerted on the lever, we need to use the equation:
Torque = Force × Distance × sin(θ)
Where:
- Torque is the rotational force (measured in Newton-meters, N·m)
- Force is the applied force (measured in Newtons, N)
- Distance is the perpendicular distance from the point of rotation to the line of action of the force (measured in meters, m)
- θ (theta) is the angle between the force and the lever arm (measured in degrees)
In this case:
- Force = 60 N
- Distance = 1.0 m
- θ = 30 degrees
First, we need to convert the angle from degrees to radians since the sine function requires radians:
θ (radians) = θ (degrees) × π/180
θ (radians) = 30 × π/180 = π/6
Now, we can calculate the torque:
Torque = Force × Distance × sin(θ)
= 60 N × 1.0 m × sin(π/6)
= 60 N × 1.0 m × 0.5
= 30 N·m
Therefore, the torque exerted on the lever is 30 N·m.