Question

\- 4.49 A large cubical block of ice of mass \(M=64 \mathrm{~kg}\) and sides of length \(L=0.40 \mathrm{~m}\) is held stationary on a frictionless ramp. The ramp is at an angle of \(\theta=26^{\circ}\) above the horizontal. The ice cube is held in place by a rope of negligible mass and length \(l=1.6 \mathrm{~m}\). The rope is attached to the surface of the ramp and to the upper edge of the ice cube, a distance \(I\) above the surface of the ramp. Find the tension in the rope.

Short answer

Answer: The tension in the rope is 627.2 N.

Step-by-step solution

Draw a Free Body Diagram (FBD)

First, let's draw a free body diagram (FBD) for the icecube. We will have the following forces acting on it: 1. Weight (W): Acts vertically downward at the center of mass of the icecube; magnitude is given by \(W=mg\), where \(m=64~kg\) is the mass of the icecube, and \(g\approx 9.8~m/s^2\) is the acceleration due to gravity. 2. Normal (N): Acts perpendicular to the ramp at the point of contact between ramp and icecube. 3. Tension (T): Acts at the upper edge of the icecube along the length of the rope that is at an angle \(\theta\) with the ramp.

Decompose Forces into Components

To analyze these forces, we need to decompose them into their components along the ramp and perpendicular to the ramp. The Weight (W) and Tension (T) can be decomposed as follows: \(W_x = W\sin\theta\) (down the ramp) \(W_y = W\cos\theta\) (perpendicular to the ramp) \(T_x = T\sin\theta\) (up the ramp) \(T_y = T\cos\theta\) (perpendicular to the ramp) Here, \(\theta=26^{\circ}\) is the given angle above the horizontal.

Apply Newton's Laws

Next, we'll apply Newton's laws along the parallel and perpendicular directions to the ramp to establish some force balance equations. Parallel to ramp: \(T_x - W_x = 0\) \(\Rightarrow T\sin\theta - W\sin\theta = 0\) Perpendicular to ramp: \(T_y + W_y -N = 0\) \(\Rightarrow T\cos\theta + W\cos\theta - N = 0\)

Simplify Equations and Solve for Tension

From the given equations in Step 3, we can simplify and solve for Tension (T). In the parallel to ramp direction: \(T\sin\theta - W\sin\theta = 0\) \(\Rightarrow T = \frac{W\sin\theta}{\sin\theta}\) \(\Rightarrow T = W\) Weight of the icecube can be calculated as: \(W = mg = 64\mathrm{~kg} \times 9.8\mathrm{~m/s^2} = 627.2\mathrm{~N}\) Therefore, tension in the rope is: \(T = W = 627.2\mathrm{~N}\)

Key concepts

Static Equilibrium

Imagine you're balancing a perfectly still object on your palm; it doesn't move. This is a classic case of static equilibrium, a fundamental concept in physics that occurs when an object is at rest or moving at a constant velocity and all the forces acting on it are perfectly balanced. In other words, not only is the net force zero, but also the net torque – the force that can cause an object to rotate – is zero.

This applies beautifully to the problem of the ice block on the ramp. Here, the ice block, although being pulled down by gravity, doesn't slide down the ramp because the forces on it are balanced. This is only possible if the net force in every direction is zero, which means the tension in the rope counters the component of the gravitational force pulling the block down the slope. Mastery of static equilibrium is essential for solving real-world engineering problems, from designing skyscrapers to ensuring the stability of bridges.

Free Body Diagram

To understand the complexity of forces at work, a free body diagram (FBD) becomes the ultimate tool for physicists and engineers. It's like a map, showing how and where forces act on a particular object. Each force is represented by a vector pointing in the direction that the force is applied.

For our chunk of ice on the inclined ramp, the FBD would have vectors for gravity pointing down, normal force pointing away from the ramp, and the tension along the rope. Such diagrams do not merely visualize, but they are also pivotal in simplifying complex systems into solvable problems by providing a clear representation of all the forces at play. Through these visual aids, one can easily decompose forces into their components and set up the equations necessary for solving physics problems.

Force Decomposition

One might wonder, how exactly do we analyze forces that aren't aligned with our conventional axes? Here's where force decomposition comes into play: it's the process of breaking a force into components that align with the axes of the system we're studying.

Let's take the gravitational force acting on the ice block. It naturally acts vertically downwards, but when placed on a slope, like in our problem, it helps to consider how much of this force pushes against the ramp (perpendicular) versus how much tries to slide the block down (parallel). By decomposing the force into these components, using simple trigonometry, it becomes much more straightforward to apply Newton's laws of motion and solve for unknowns like the tension in a rope or the normal force exerted by a surface.

Newton's Laws of Motion

Newton's laws of motion are the ABCs of physics, the foundational principles that describe how objects move. There are three laws: the first is about inertia and states that objects will remain in their state of motion unless a force acts on them. The second provides the famous equation \( F=ma \), which relates force, mass, and acceleration. The third is about action and reaction, stating that forces always come in pairs with equal magnitude and opposite direction.

In terms of our ice block problem, Newton's laws help us understand the motion (or lack thereof) of the block. The second law, in particular, tells us that because the block is in static equilibrium (it’s not accelerating), the net force acting on it must be zero. By applying this law, we can set up equations that help us determine the tension in the rope keeping the block stationary on the ramp.