A 20.5 N weight is held in the outstretched hand of a person with a shoulder-to-elbow length of 30.0 cm and an elbow-to-wrist length of 24.0 cm. The center of the weight is 5.5 cm from the person's wrist. Determine the magnitude of the torque about the elbow produced by the weight.
Oh, the joy of torque calculations! Let's get to it, shall we?
To calculate the torque (τ) around the elbow, we need to multiply the force (F) by the perpendicular distance from the axis of rotation (the elbow) to the line of action of the force.
So, in this case, the force (F) is 20.5 N, and the distance from the elbow (axis of rotation) to the line of action of the force is the sum of the shoulder-to-elbow length (30.0 cm) and the elbow-to-wrist length (24.0 cm), which gives us a total of 54.0 cm or 0.54 m.
But wait, we still need to account for the slight offset of 5.5 cm from the person's wrist to the center of the weight. So, we need to subtract that offset from the total distance, which gives us 0.54 m - 0.055 m = 0.485 m.
Now that we have both the force (F = 20.5 N) and the distance (d = 0.485 m), we can calculate the torque (τ) using the formula: τ = F * d.
Plugging in the values, we get τ = 20.5 N * 0.485 m = 9.9425 Nm.
So, the magnitude of the torque about the elbow produced by the weight is approximately 9.9425 Newton-meters. And just like that, we've unlocked the secrets of torque!
To determine the magnitude of the torque about the elbow produced by the weight, we need to calculate the torque due to the force exerted by the weight.
Torque (τ) is defined as the product of the force (F) and the perpendicular distance (r) from the pivot point (in this case, the elbow) to the line of action of the force. The equation for torque is:
τ = F * r * sin(θ)
Where:
τ = torque
F = force
r = perpendicular distance
θ = angle between the force vector and the line connecting the pivot point and the point of application of the force
In this case, the force exerted by the weight is 20.5 N and the perpendicular distance from the weight's center to the person's wrist is 5.5 cm = 0.055 m.
The angle θ is the angle between the force vector and the line connecting the pivot point (elbow) and the point of application of the force (wrist). Since the weight is held in an outstretched hand, the angle θ is 180 degrees (or π radians), as the force vector and the line connecting the pivot and point of application are in opposite directions.
Substituting the given values into the torque equation:
τ = 20.5 N * 0.055 m * sin(π)
Since sin(π) is equal to 0, the torque is equal to 0.
Therefore, the magnitude of the torque about the elbow produced by the weight is 0 Nm.
To determine the magnitude of the torque about the elbow produced by the weight, we can use the formula for torque:
Torque = Force x Lever arm
In this case, the force is the weight of 20.5 N and the lever arm is the distance between the weight's center and the elbow. Let's calculate the lever arm:
Lever arm = Shoulder-to-elbow length + Elbow-to-wrist length - Distance from wrist to weight's center
Lever arm = 30.0 cm + 24.0 cm - 5.5 cm
Lever arm = 48.5 cm
Now, we can calculate the torque:
Torque = Force x Lever arm
Torque = 20.5 N x 48.5 cm
Now, we need to convert cm to meters to keep the units consistent:
Torque = 20.5 N x (48.5 cm / 100 cm)
Torque = 20.5 N x 0.485 m
Torque = 9.9325 N*m
Therefore, the magnitude of the torque about the elbow produced by the weight is approximately 9.9325 N*m.