Question

On the bunny hill at a ski resort, a towrope pulls the skiers up the hill with constant speed of \(1.74 \mathrm{~m} / \mathrm{s}\). The slope of the hill is \(12.4^{\circ}\) with respect to the horizontal. A child is being pulled up the hill. The coefficients of static and kinetic friction between the child's skis and the snow are 0.152 and 0.104 , respectively, and the child's mass is \(62.4 \mathrm{~kg}\), including the clothing and equipment. What is the force with which the towrope has to pull on the child?

Short answer

Answer: The force with which the towrope has to pull on the child is 195.1 N.

Step-by-step solution

Calculate gravitational force

To calculate the gravitational force acting on the child, we use the equation \(F_g = m \cdot g\), where \(F_g\) is the gravitational force, \(m = 62.4\mathrm{~kg}\) is the mass of the child, and \(g = 9.81 \mathrm{~m/s^2}\) is the acceleration due to gravity. \(F_g = 62.4 \mathrm{~kg} \cdot 9.81 \mathrm{~m/s^2} = 611.9 \mathrm{~N}\)

Resolve gravitational force into horizontal and vertical components

To analyze the forces separately, we need to find the components of the gravitational force in the horizontal and vertical directions. We can use the given angle of the slope, \(12.4^{\circ}\), to split the gravitational force into components. Vertical component: \(F_{g_{v}} = F_g \cdot \cos(12.4^{\circ})\) Horizontal component: \(F_{g_{h}} = F_g \cdot \sin(12.4^{\circ})\) \(F_{g_{v}} = 611.9 \mathrm{~N} \cdot \cos(12.4^{\circ}) = 597.4 \mathrm{~N}\) \(F_{g_{h}} = 611.9 \mathrm{~N} \cdot \sin(12.4^{\circ}) = 133.0 \mathrm{~N}\)

Calculate frictional force

The frictional force acting on the child is given by \(F_f = \mu \cdot F_{g_{v}}\), where \(F_f\) is the frictional force, \(\mu = 0.104\) is the coefficient of kinetic friction, and \(F_{g_{v}}\) is the vertical component of the gravitational force. \(F_f = 0.104 \cdot 597.4 \mathrm{~N} = 62.1 \mathrm{~N}\)

Apply Newton's second law

According to Newton's second law, the net force acting on the child must equal the child's mass times acceleration in both vertical and horizontal directions. In the horizontal direction, we have \(F_{tow} - F_{g_{h}} - F_f = m \cdot a\), where \(F_{tow}\) is the force exerted by the towrope and \(a = 0 \mathrm{~m/s^2}\) (since the child is moving with constant speed). By rearranging the equation, \(F_{tow} = F_{g_{h}} + F_f\). Substituting the calculated values, we get \(F_{tow} = 133.0 \mathrm{~N} + 62.1 \mathrm{~N} = 195.1 \mathrm{~N}\) So the force with which the towrope has to pull on the child is \(195.1 \mathrm{~N}\).

Key concepts

Gravitational Force

Gravitational force is a fundamental force that attracts any two objects with mass. Every object, including skiers on a hill, experiences this force due to their mass and the Earth's mass.

In physics problems, the gravitational force (\(F_g\)) is often simplified to the weight of the object and is calculated using the equation \(F_g = m \times g\), where \(m\) is the object's mass and \(g\) is the acceleration due to gravity, approximately \(9.81 \, \mathrm{m/s^2}\) on Earth. For a person skiing, it's crucial to understand how gravitational force affects their motion down the slopes.

While this force acts directly downward, its impact changes if the individual is on a hill at an angle, like the bunny hill in the exercise. The force can be resolved into two components: one perpendicular to the hill's surface and one parallel to it. The component parallel to the slope is the driving force that causes the skier to move downhill.

Frictional Force

Frictional force acts opposite to the direction of motion, resisting the sliding motion between two surfaces. When it comes to skiing, frictional force is between the skis and the snow, and it's crucial for controlling speed and for stopping.

The force of friction (\(F_f\)) depends on the normal force - the force exerted by a surface perpendicular to the object, and the coefficient of friction (\(\mu\)). There are two types of frictional coefficients: static (\(\mu_s\)) and kinetic (\(\mu_k\)). Static friction keeps objects at rest while kinetic friction acts on moving objects.

In the given problem, the relevant coefficient is kinetic because the skier is already in motion. Here, the frictional force is calculated as \(F_f = \mu_k \cdot F_{g_v}\), where \(F_{g_v}\) is the vertical component of the gravitational force facing the hill. Remember that friction doesn't just stop movements; it also allows the skier to maintain control and affects the tension in the towrope.

Newton's Second Law

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (\(F = m \times a\)). In simpler terms, this law describes how an object will move when forces are applied to it.

In the context of the exercise, the skier is moving at a constant speed, meaning that the acceleration (\(a\)) is zero. When acceleration is zero, it implies that the net force in the direction of motion is also zero, so all the forces are balanced. The forces in question here include the gravitational force parallel to the slope, the frictional force, and the tension in the towrope.

A skier being pulled up a hill by a towrope is a classic example of this law in action. The towrope must exert a force that balances the components of gravitational force pulling the child down the slope and the frictional force resisting the motion. This equilibrium allows the skier to ascend the hill at a constant velocity.