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###### Torque

Physics (Year 12) - Gravity and Motion

Dev Lohar

# What is Torque?

In formal terms, torque (also called moments) is the measure of force required to rotate an object about an axis. The axis is usually called the pivot point or axis of rotation. For example, the simple act of opening or closing a door is related to torque. Relating this example to the definition, the object is the door and its rotating around its hinge; the pivot point. Lifting a dumbbell is related to torque as well. The object is your arm and the pivot point is the elbow.

To provide torque, a force must be applied that is not aligned with the pivot point. Implying that there must be some distance from the line of action of the force and the pivot point. For example, you cannot open a door by pushing at the hinge. You have to push at a distance from the hinge.

Consider the wheel below shown at 2 angles. The pivot point of the wheel is the centre and is the axis about which the wheel rotates. Applying a force at the centre would not cause the wheel to experience a torque. But if we apply a force at the edge of wheel, torque will be achieved because the line of action (a line in the same direction as the applied force) does not go through the pivot point. If it does, no torque is achieved. Maximum torque is achieved when the force is 90o to the pivot point like in the second diagram below.

# Torque Equation

The equation to calculate torque is:

Through the equation we can see that the larger the force or the perpendicular distance, the larger the torque. We also notice that to provide the same amount of torque, a larger perpendicular distance can be used but with a smaller force. For example, a longer spanner will require less force to be applied to unscrew a hex nut compared to a shorter spanner which will require more force to provide the same amount of torque.

# Non-Perpendicular Calculations

The torque equation is easy to use if the line of action is perpendicular to the axis of rotation like in the wheel diagram above. But most of the times, there are scenarios where the force causing torque is acting along a line that is not at 90 degrees to the pivot point axis. In such cases the torque is reduced and 2 different methods can used to calculate it. It can be calculated by using the perpendicular force, or by using the perpendicular distance. However, both methods utilise the same equation:

When calculating torque using perpendicular force, we use the distance from the pivot point to the line of action as it's given in the question, but use the perpendicular force. We find the component of the force which is perpendicular to the pivot point, as seen in the diagram below. Given we know the angle between the line of action and the lever arm, we can use FsinÎ¸ to find the perpendicular force. Now we just multiply the perpendicular force with the distance to give us torque; Ï„ = rFsinÎ¸.

When calculating torque using perpendicular distance, we use the force as its given in the question, but use the perpendicular distance. We find the perpendicular distance from the pivot point to the line of action as seen in the diagram below. This can be calculated by rsinÎ¸ where we know the angle between the line of action and the lever arm. Now we just multiply the perpendicular distance with the force to give us the torque; Ï„ = rFsinÎ¸.

# Direction of torque

Torque is a vector, meaning it has both magnitude and direction. The most common directions used for torque is clockwise and anticlockwise. Consider the diagram below of a balanced seesaw with a pivot point in the middle.

If we now place a 30kg child 1 metre to the left of the pivot point, it would cause the seesaw to rotate about its pivot point and experience torque. To find the magnitude of the torque we use Ï„=rF where r is 1 metre and F is the force of gravity which is 30 x 9.8 = 294, hence the torque is 1 x 294 = 294 Nm. To find the direction we need to focus on which side of the pivot point the force is applied (i.e. to the left or to the right) and if the force is acting up or down. In this example, the force (which is the force of gravity) is to the left of the pivot point and acting down. Now imagine drawing a circle from the end of the force vector towards the pivot point. Which direction would we begin to draw the circle? That will be the direction of the torque. In this example, we would go to the right and up, hence weâ€™d be moving in an anticlockwise direction as seen in the diagram below.

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