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

Answer: The magnitude of the tangential force needed to hold the ice block in place is 455.18 N, and the magnitude of the horizontal force needed to hold the ice block in place is 784.80 N.

Step-by-step solution

Identify the forces acting on the ice block

The forces acting on the block are: 1. Gravitational force, acting vertically downward with magnitude \(Mg\). 2. Tangential force, acting at an angle of \(\theta\) above the horizontal.

Resolve the gravitational force into components

We need to find the components of gravitational force parallel and perpendicular to the ramp. The parallel component is \(Mg \sin\theta\), and the perpendicular component is \(Mg \cos\theta\).

Apply Newton's second law of motion

The ice block is in equilibrium along both the parallel and perpendicular directions of the ramp, so the net force acting along both directions is zero. For the parallel direction: \(F_{tangential} - Mg \sin\theta = 0\)

Solve for the tangential force

Rearrange the equation from Step 3 to obtain the tangential force: \(F_{tangential} = Mg \sin\theta\) Plug in the given values of mass M=80 kg, angle \(\theta = 36.9^\circ\), and gravitational acceleration \(g = 9.81 \mathrm{~m/s^2}\): \(F_{tangential} = 80 \times 9.81 \times \sin(36.9^\circ) = 455.18 \mathrm{~N}\) So, the magnitude of the tangential force needed to hold the ice block in place is 455.18 N. #b) Horizontal Force Directed Toward the Center of the Ice Block#

Identify the forces acting on the ice block

The forces acting on the block are: 1. Gravitational force, acting vertically downward with magnitude \(Mg\). 2. Horizontal force, directed horizontally toward the center of the ice block.

Resolve the gravitational force into components

We will again resolve the gravitational force into components parallel and perpendicular to the ramp. The parallel component is \(Mg \sin\theta\), and the perpendicular component is \(Mg \cos\theta\).

Apply Newton's second law of motion

As the ice block is in equilibrium along both the parallel and perpendicular directions of the ramp, the net force acting along both directions is zero. For the parallel direction: \(F_{horizontal} \sin\theta - Mg \sin\theta = 0\) For the perpendicular direction: \(F_{horizontal} \cos\theta - Mg \cos\theta = 0\)

Solve for the horizontal force

Using the equation for the parallel direction to solve for the horizontal force: \(F_{horizontal} \sin\theta = Mg \sin\theta\) Rearranging, we get: \(F_{horizontal} = \frac{Mg \sin\theta}{\sin\theta}\) Plug in the given values of mass M=80 kg, angle \(\theta = 36.9^\circ\), and gravitational acceleration \(g = 9.81 \mathrm{~m/s^2}\): \(F_{horizontal} = \frac{80 \times 9.81 \times \sin(36.9^\circ)}{\sin(36.9^\circ)} = 784.80 \mathrm{~N}\) So, the magnitude of the horizontal force needed to hold the ice block in place is 784.80 N.

Key concepts

Newton's Second Law of Motion

Understanding Newton's second law of motion is crucial when dealing with physics equilibrium problems. This law can be succinctly stated as F=ma, where F represents force, m is mass, and a is acceleration. The beauty of this law lies in its applicability; it allows us to analyze how objects will move under the influence of various forces.

When an object is in equilibrium, as the ice block in our exercise, its acceleration is zero. Accordingly, the second law simplifies to the statement that the sum of all forces acting upon the object must be zero. In essence, all forces balance each other out. For the ice block on an inclined plane, there are two scenarios: it's either held by a tangential force along the ramp or by a horizontal force. In both cases, Newton's second law enables us to establish relationships between these forces to maintain the equilibrium – no net force means no acceleration, and thus the block remains stationary.

Gravitational Force Components

The concept of gravitational force components is pivotal when dissecting the conditions of equilibrium, particularly on an inclined plane. Gravitational force, commonly symbolized as Mg where M is mass and g is the acceleration due to gravity, acts straight downwards towards the center of the Earth. However, when on an incline, this gravitational pull doesn't align with the direction of potential movement of the object.

Therefore, we decompose this force into two components: one parallel to the plane, Mg sin(θ), and another perpendicular to it, Mg cos(θ). This simplification into components allows us to analyze the effects of gravity in the distinct directions defined by the geometry of the object's environment – the inclined ramp in our exercise. It's these components that play a direct role in determining the magnitude of the forces required to keep our ice block stationary.

Equilibrium on an Inclined Plane

Tackling equilibrium on an inclined plane exercises involves a marriage of concepts: Newton's second law of motion and the understanding of gravitational force components. An incline adds complexity because it introduces a new angle into considerations, changing how the force vectors work.

For an object to remain in equilibrium on an incline, the sum of forces along the slope (tangential forces) and perpendicular to the slope must be zero. This means that for the object not to accelerate along the plane, any component of the gravitational force running parallel to the incline must be counteracted by a force of equal magnitude but in the opposite direction. An understanding of the angle of the incline is vital here, as it influences the magnitude of the force necessary to achieve equilibrium, as demonstrated through our calculation of the tangential force.

Similarly, replacement of this tangential force with a horizontal force introduces a different scenario while still adhering to equilibrium principles. By analyzing the gravitational components, we deduce that the horizontal force must also be adjusted to address the inclination angle, resulting in a uniquely different force requirement than that of the tangential force to maintain a stationary state for the ice block.