Question

A large ice block of mass \(M=80.0 \mathrm{~kg}\) is held stationary on a frictionless ramp. The ramp is at an angle of \(\theta=36.9^{\circ}\) above the horizontal. a) If the ice block is held in place by a tangential force along the surface of the ramp (at angle \(\theta\) above the horizontal), find the magnitude of this force. b) If, instead, the ice block is held in place by a horizontal force, directed horizontally toward the center of the ice block, find the magnitude of this force.

Short answer

Question: Calculate the magnitudes of the forces holding the ice block in place in the two different scenarios: (a) A force tangential to the ramp's surface. (b) A horizontal force directed toward the center of the ice block. Answer: (a) The magnitude of the tangential force is approximately 466.4 N. (b) The magnitude of the horizontal force is approximately 601.9 N.

Step-by-step solution

Resolve the gravitational force into components

The gravitational force acting on the ice block has a magnitude of \(Mg\), where \(g\) is the acceleration due to gravity. This force can be resolved into two components: one parallel to the ramp's surface and one perpendicular to it. The component parallel to the ramp can be found using the equation \(F_{parallel} = Mg\sin(\theta)\) and the perpendicular component using the equation \(F_{perpendicular} = Mg\cos(\theta)\).

Apply Newton's second law

Since the ice block is stationary and there is no friction, the net force acting on it is zero. In the parallel direction, the tangential force \(F_{tangential}\) must balance the gravitational force component. Thus, we can write \(F_{tangential} = F_{parallel}\).

Calculate the tangential force

Now, we substitute the expression for \(F_{parallel}\) from step 1 into the equation from step 2 and solve for the tangential force: \[F_{tangential} = Mg\sin(\theta)\] Plug in the given values of \(M = 80.0\,\text{kg}\), \(g = 9.8\,\text{m/s²}\), and \(\theta = 36.9^\circ\), and calculate the magnitude of the tangential force: \[F_{tangential} = 80.0\,\text{kg} \cdot 9.8\,\text{m/s²} \cdot \sin(36.9^\circ) \approx 466.4\,\text{N}\] #b) Horizontal force toward the center#

Resolve the gravitational force into components

Just as in part (a), we resolve the gravitational force into components parallel and perpendicular to the ramp's surface using the same equations.

Apply Newton's second law

In this case, we have a horizontal force \(F_{horizontal}\) that balances the gravitational force component in the horizontal direction. Since the ramp is frictionless, we only need to consider the component of gravitational force parallel to the ramp. Thus, we can write \(F_{horizontal} \cos(\theta) = F_{parallel}\).

Calculate the horizontal force

Now, we substitute the expression for \(F_{parallel}\) from step 1 into the equation from step 2, and solve for the horizontal force: \[F_{horizontal} \cos(\theta) = Mg\sin(\theta)\] \[F_{horizontal} = \frac{Mg\sin(\theta)}{\cos(\theta)}\] Plug in the given values of \(M = 80.0\,\text{kg}\), \(g = 9.8\,\text{m/s²}\), and \(\theta = 36.9^\circ\), and calculate the magnitude of the horizontal force: \[F_{horizontal} = \frac{80.0\,\text{kg} \cdot 9.8\,\text{m/s²} \cdot \sin(36.9^\circ)}{\cos(36.9^\circ)} \approx 601.9\,\text{N}\] In conclusion, when the ice block is held in place by a tangential force along the surface, the magnitude of the force is approximately \(466.4\,\text{N}\), and when it is held in place by a horizontal force towards its center, the magnitude of the force is approximately \(601.9\,\text{N}\).

Key concepts

Gravitational Force Components

Understanding the components of gravitational force is crucial when dealing with physics problems on an incline, such as the scenario with the ice block on a frictionless ramp. In physics, weight, or gravitational force, always acts directly downwards towards the center of the Earth, which we represent with the acceleration constant, \( g \), typically approximated as \( 9.8\text{ m/s}^2 \).

When an object is placed on an angled surface, this force doesn't disappear but rather gets divided into two perpendicular components: one pushing into the incline (perpendicular) and one running parallel down the incline (parallel).

The mathematical expressions for these components are simple trigonometric functions of the angle of incline, \( \theta \). The component parallel to the incline is obtained by multiplying the total weight by the sine of the angle, \( Mg\sin(\theta) \), and is responsible for the object's tendency to slide down the incline. Conversely, the component perpendicular to the incline, \( Mg\cos(\theta) \), pushes the object into the surface and is a critical factor when considering normal force and potential frictional forces.

This resolution of forces is the foundation for analyzing forces on inclines and is key in many physics problems.

Newton's Second Law

Newton's second law of motion tells us how force, mass, and acceleration are related. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, expressed as \( F = ma \).

For stationary objects, like our ice block example, Newton's second law implies that all the forces acting on the body must be in equilibrium, which means they cancel each other out resulting in no net force, and thus no acceleration. This is why the force required to keep the ice block stationary on the ramp must counteract exactly the downhill gravitational component, showing that forces do not necessarily cause motion, but rather changes in motion.

When applying Newton's second law to problems involving inclined planes, remember that you're working with the components of the gravitational force. It is the parallel component that has the potential to cause acceleration down the plane, hence why we balance it with an external force to keep the object stationary, as seen with the tangential and horizontal forces in the ice block scenario.

Frictionless Incline Problem

An idealized frictionless incline is a common concept in physics problems to simplify calculations by ignoring the effects of friction. Friction is the resistive force that opposes motion; without it, any component of force parallel to the incline will cause the object to accelerate along that surface.

In the context of our ice block scenario, by assuming a frictionless plane, we focus solely on the balance of gravitational force components and the forces applied to keep the block stationary. This simplifies our calculations to finding the equivalent opposing force that would counteract the gravitational pull down the ramp.

A frictionless incline problem not only sharpens our understanding of the resolution of forces and application of Newton's second law but also provides an accessible entry point to more complex scenarios. As students master the frictionless problem, they can then add layers of complexity, such as friction or external applied forces, to model real-world situations more accurately. However, the core principles established in the frictionless case remain vital as foundational knowledge.