Problem 99
Question
Block \(A,\) with weight \(3 w,\) slides down an inclined plane \(S\) of slope angle \(36.9^{\circ}\) at a constant speed while plank \(B\) with weight \(w,\) rests on top of A. The plank is attached by a cord to the wall (Fig. P5.99). (a) Draw a diagram of all the forces acting on block A. (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A,\) determine its value.
Step-by-Step Solution
Verified Answer
The coefficient of kinetic friction is approximately 0.33.
1Step 1: Understand the Problem Setup
Block A slides down an inclined plane with a constant speed, which means the net force acting along the plane is zero. Block B rests on top of Block A and is held in place by a cord attached to a wall. This tells us that static and kinetic friction play essential roles in maintaining the system's dynamics.
2Step 2: Identify the Forces on Block A
For Block A, identify the forces: weight of Block A (\[3w\], acting vertically downwards), normal force (\[N_A\], perpendicular to the incline), gravitational component parallel to the incline (\[3w \sin(36.9^\circ)\]), kinetic friction opposing motion (\[f_k\]), and tension from Block B inhibited by the cord.
3Step 3: Calculate Forces Perpendicular to Plane
Calculate the normal force \(N_A\) acting on Block A using the resolved component of gravity perpendicular to the incline.\[\begin{align*}N_A &= 3w \cos(36.9^\circ)\end{align*}\]
4Step 4: Calculate Forces Parallel to Plane
The force due to gravity parallel to the incline is balanced by the frictional force.\[\begin{align*}3w \sin(36.9^\circ) &= f_k + w\end{align*}\]Where kinetic friction \(f_k\) is \(\mu_k N_A\).
5Step 5: Express Friction in Terms of Kinetic Coefficient
Substitute \(f_k = \mu_k N_A\) into the previous equation:\[\begin{align*}3w \sin(36.9^\circ) &= \mu_k \cdot 3w \cos(36.9^\circ) + w\end{align*}\]This relationship allows us to solve for \(\mu_k\).
6Step 6: Solve for the Coefficient of Kinetic Friction
Rearrange the equation from Step 5 to solve for \(\mu_k\):\[\begin{align*}3 \sin(36.9^\circ) - 1 &= \mu_k \cdot 3 \cos(36.9^\circ) \\mu_k &= \frac{3 \sin(36.9^\circ) - 1}{3 \cos(36.9^\circ)}\end{align*}\]
7Step 7: Calculate and Finalize the Coefficient Value
Insert the trigonometric values \(\sin(36.9^\circ) = 0.6\) and \(\cos(36.9^\circ) = 0.8\) into the equation:\[\begin{align*}\mu_k &= \frac{3 \, \times \, 0.6 - 1}{3 \, \times \, 0.8} \\mu_k &= \frac{1.8 - 1}{2.4} \\mu_k &= \frac{0.8}{2.4} \\mu_k &= \frac{1}{3}\end{align*}\]Thus, the coefficient of kinetic friction is approximately 0.33.
Key Concepts
Inclined Plane PhysicsKinetic Friction CoefficientNewton's Laws of Motion
Inclined Plane Physics
Inclined plane physics is a fascinating area of study that explores how objects behave when placed on slopes. An inclined plane, like the one mentioned in the problem, is simply a flat surface tilted at an angle compared to the horizontal ground. This setup allows us to see forces at work, particularly gravity and friction. When a block is placed on such an incline, gravity helps to pull it down the slope.
The slope's angle is crucial because it affects how the forces are split into parallel and perpendicular components relative to the plane. For example, in our exercise, the angle of the incline is 36.9°. This angle impacts how much of the gravitational force attempts to slide the block down, and how much presses against the plane. The gravitational component pulling the block downward is calculated using \(3w \sin(36.9^\circ)\), and the component pushing the block into the incline is \(3w \cos(36.9^\circ)\).
This scenario is a great opportunity to observe Newton's laws of motion at work, as the forces must balance for the block to slide at a constant speed.
The slope's angle is crucial because it affects how the forces are split into parallel and perpendicular components relative to the plane. For example, in our exercise, the angle of the incline is 36.9°. This angle impacts how much of the gravitational force attempts to slide the block down, and how much presses against the plane. The gravitational component pulling the block downward is calculated using \(3w \sin(36.9^\circ)\), and the component pushing the block into the incline is \(3w \cos(36.9^\circ)\).
This scenario is a great opportunity to observe Newton's laws of motion at work, as the forces must balance for the block to slide at a constant speed.
Kinetic Friction Coefficient
Friction is a force that resists the motion of objects sliding against each other. The kinetic friction coefficient, denoted by \(\mu_k\), quantifies this resistance when the surfaces are in motion. In the context of our problem, friction is essential as it keeps block A moving at a constant speed down the incline. Frictional force works against the gravitational pull, ensuring there's no acceleration.
The kinetic friction force \(f_k\) is calculated using the formula: \(f_k = \mu_k N_A\), where \(N_A\) is the normal force pressing the object into the plane. This frictional force modifies the net movement of block A, so it doesn't accelerate. By balancing the forces parallel to the plane, we can solve for the kinetic friction coefficient. Using the given values and calculations, \(\mu_k\) is determined to be approximately 0.33.
The kinetic friction force \(f_k\) is calculated using the formula: \(f_k = \mu_k N_A\), where \(N_A\) is the normal force pressing the object into the plane. This frictional force modifies the net movement of block A, so it doesn't accelerate. By balancing the forces parallel to the plane, we can solve for the kinetic friction coefficient. Using the given values and calculations, \(\mu_k\) is determined to be approximately 0.33.
- The normal force varies based on the block’s weight and the incline angle.
- The friction coefficient depends on the materials' surfaces in contact.
Newton's Laws of Motion
Newton's laws of motion are foundational principles that describe the behavior of objects in motion. Let's unpack how they apply to our inclined plane scenario.
- First Law (Law of Inertia): An object in motion stays in motion with the same speed and in the same direction unless acted upon by a net external force. In this problem, block A slides down at a constant speed, indicating balanced forces.
- Second Law (Law of Acceleration): The acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass. Here, since Block A moves at a constant speed, the net force parallel to the incline is zero \((F = ma = 0)\). This confirms the balance between the gravitational component down the slope and the kinetic frictional force.
- Third Law (Action and Reaction): For every action, there is an equal and opposite reaction. The normal force exerted by the incline on the block is equal and opposite to the gravitational force component pressing against the hill.
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