I need to explain and demonstrate mathematically why it is easier to pull a wheelbarrow than it is to push it.
Also, what effect does the component of force parallel to a bicycle crank have?
To explain mathematically why it is easier to pull a wheelbarrow than to push it, we can consider the forces involved and analyze the concept of friction.
When you push a wheelbarrow, you are applying a force in a forward direction parallel to the ground, which we will call the applied force (F_app). Simultaneously, the wheelbarrow exerts an equal and opposite force on the ground, called the friction force (F_friction). This friction force opposes the motion and makes it harder to push the wheelbarrow.
Now let's compare this to pulling the wheelbarrow. When you pull the wheelbarrow, the applied force is still in the forward direction parallel to the ground (F_app). However, the difference lies in the direction of the friction force. In this case, the friction force is applied in the same direction as the applied force, aiding the motion instead of opposing it.
We can represent this mathematically using free-body diagrams. Suppose the effective force or effort you apply is represented by F_app, and the coefficient of friction between the wheelbarrow and the ground is represented by μ. The friction force (F_friction) can be calculated using the equation: F_friction = μ * normal force, where the normal force is the force exerted by the ground on the wheelbarrow, typically equal to its weight.
When pushing the wheelbarrow, the friction force opposes the motion and is equal to F_friction = μ * normal force. Whereas, when pulling the wheelbarrow, the friction force aids the motion and is equal to F_friction = -μ * normal force (negative sign indicates the direction of the force).
So mathematically, it is easier to pull the wheelbarrow because the forces act in the same direction, making the net effective force (F_app + F_friction) larger in magnitude and making it easier to overcome the friction force.
Regarding the effect of the component of force parallel to a bicycle crank, it plays a crucial role in propelling the bicycle forward. This force, often referred to as the tangential force, is created by applying a force on the pedals parallel to the crank arm. The tangential force is responsible for rotating the crank, which in turn drives the chain and the rear wheel, ultimately propelling the bicycle.
The magnitude of the tangential force acting on the bicycle crank is given by the equation: F_tangential = F_applied * R, where F_applied is the force applied on the pedal, and R is the distance from the center of the crank to the axis of rotation.
The effect of the tangential force is directly related to the torque it generates. The torque is the product of the tangential force and the distance between the axis of rotation and the line of action of the force. For example, when the pedal is at a perpendicular distance R from the axis of rotation (crank arm length), the torque is maximum, resulting in maximum propulsion. As the pedal moves closer to the axis of rotation, the torque decreases, reducing the effectiveness of the force in propelling the bicycle.
In summary, the component of force parallel to a bicycle crank is essential for generating torque, which drives the bicycle forward. By applying force tangentially to the crank, riders can efficiently convert their applied force into rotational motion, propelling the bicycle.