The handle of a 22 -kg lawnmower makes a \(35^{\circ}\) angle with the horizontal. If the coefficient of friction between lawnmower and ground is \(0.68,\) what magnitude of force, applied in the direction of the handle, is required to push the mower at constant velocity? Compare with the mower's weight.

The coefficient of friction is a crucial value in physics that characterizes the amount of frictional resistance between two surfaces. In the context of the lawnmower problem, it's represented by the symbol \(\mu\) and is given as 0.68. This number essentially tells us how 'sticky' or 'slippery' the surface contact is between the lawnmower's wheels and the ground. A higher coefficient indicates more friction, thus requiring more force to move the object at a constant velocity. To maintain a steady pace without acceleration, the force applied to the mower must be equal to the frictional resistance. This reliance on the coefficient of friction is integral to calculating the precise force needed to push the mower across the yard.

Achieving a constant velocity with the lawnmower means that it's moving without speeding up or slowing down. In physics, this implies that the net force acting on the lawnmower is zero because force is directly linked to acceleration according to Newton's second law. When we apply force to the mower in the direction of the handle, we aim to overcome friction and any other forces that might be decelerating it, such as wind resistance or a slight incline.

To maintain a constant velocity, the horizontal component of the applied force must exactly counterbalance the frictional force working against it. This delicate balancing act is what enables the user to push the lawnmower smoothly without applying varying force levels, thus ensuring that the neatness of the lawn cut isn't compromised by changes in the speed of the mower.

The normal force is the supportive force exerted by a surface to support the weight of an object placed on it; think of it as the ground pushing back against the lawnmower with a force equal to its weight when there's no additional vertical force applied. This force is 'normal' in the sense that it is perpendicular to the contact surface.

In the exercise, the normal force plays a pivotal role since it affects the frictional force. Because the handle is angled, part of the applied force actually reduces the normal force the ground exerts on the mower by lifting it slightly. This reduction in normal force, in turn, decreases the frictional force because friction depends on the normal force. For the mower's constant velocity movement, accurately determining the normal force is essential as it directly correlates to how much friction needs to be overcome.

The acceleration due to gravity, often symbolized as \(g\), is the rate at which objects accelerate towards the Earth as a result of its gravitational pull. Standardized as approximately \(9.8\;m/s^2\), this acceleration impacts all objects on Earth's surface. In calculating the lawnmower's force requirements, gravity's role is twofold: it determines the weight of the lawnmower, and it acts as a constant factor in computations involving the normal and frictional forces.

Understanding the acceleration due to gravity is critical as it helps to establish the baseline forces at play before any additional effort is applied. It allows us to calculate the lawnmower’s weight, which is needed to subsequently derive the normal force and the frictional force. Comprehending how gravity influences these elements aids students in grasping the relationships among the various forces involved in simple mechanical tasks, like pushing a lawnmower.