Problem 39 - An Introduction to the Mechanics of Solids in SI Units

Question

A block of weight \(W\) rests on an inclined plane which makes an angle \(\theta=\tan ^{-1} 3 / 4\) as shown. A force \(P\), parallel to the \(x\) axis, is applied to the block and gradually increased from zero; when \(P\) reaches the value \(0.4 W\) the block begins to slide. What is the coefficient of friction between the block and the inclined plane?

Step-by-Step Solution

Answer

The coefficient of friction between the block and the inclined plane is 1.

Step 1: Identify the forces acting on the block

There are three main forces acting on the block: the weight \(W\) acting vertically downwards, the normal force \(N\) acting perpendicular to the inclined plane, and the applied force \(P\) acting horizontally. The force of static friction \(f_s\) acts parallel to the plane, opposing the tendency of the block to slide.

Step 2: Resolve the weight into components

The weight \(W\) of the block can be split into two components: one parallel to the inclined plane (\(W\sin(\theta)\)) and the other perpendicular to the plane (\(W\cos(\theta)\)). Since the plane is inclined at an angle \(\theta = \tan^{-1}(3/4)\), the components are \(W_\parallel = W\sin(\theta) = W \cdot (3/5)\) and \(W_\perp = W\cos(\theta) = W \cdot (4/5)\).

Step 3: Calculate the normal force

The normal force \(N\) is the force exerted by the inclined plane on the block in the direction perpendicular to the plane. It cancels out the component of the weight perpendicular to the plane. Therefore, \(N = W_\perp = W \cdot (4/5)\).

Step 4: Determine when the block starts to slide

The block begins to slide when the force applied on it is equal to the force of static friction. The force of static friction is given by \(f_s = \mu_s N\), where \(\mu_s\) is the coefficient of static friction. We also know that the applied force \(P = 0.4W\) when the block starts to slide. Moreover, the component of weight parallel to the plane adds to the force tending to slide the block. Setting the sum of forces parallel to the plane to zero gives \(W_\parallel + \mu_s N = P\). Substituting the known values gives \(W \cdot (3/5) + \mu_s W \cdot (4/5) = 0.4W\).

Step 5: Calculate the coefficient of static friction

Solve the previous equation for \(\mu_s\). This gives \(\mu_s = (0.4 - 0.6) / 0.8 = -0.25\). Since the coefficient of friction cannot be negative, we have made a mistake. Considering the forces properly, we realize that when the block is about to slide, the frictional force \(f_s\) should balance not just the component of the weight \(W_\parallel\) but also the applied force \(P\). The corrected equation is \(\mu_s N = W_\parallel + P\), from which it gives \(\mu_s = (W \cdot (3/5) + 0.4W) / (W \cdot (4/5)) = 1\).

Step 6: Interpret the result

The coefficient of friction is a dimensionless quantity that describes the ratio of the force of friction between two bodies to the force pressing them together. A coefficient of 1 means the frictional force is equal to the normal force. In this case, it is required to overcome a force equal to the weight of the block in order to make it slide along the plane. The force applied has to balance both the component of weight parallel to the plane and the frictional force which is as large as the normal force.

Key Concepts

Inclined Plane Problem

Inclined plane problems are typical in physics and engineering as they embody a host of forces and principles at work. An inclined plane, simply put, is a flat surface tilted at an angle to the horizontal. These problems become an excellent practical application of the concepts of static friction, normal force, and mechanics of solids.
They show how different forces interact and affect the motion of an object. In our problem, the inclined plane creates a scenario where gravity does not act perpendicular to the surface but along a tilted path, introducing the necessity to decompose forces into parallel and perpendicular components.

Decomposing Forces on an Inclined Plane

• The weight of the object is split into a component parallel to the plane, which contributes to potential sliding, and a component perpendicular to the plane, which contributes to the normal force.
• Static friction adjusts to counterbalance any parallel forces up to its maximum limit, which is governed by the normal force and the coefficient of static friction.

This example illustrates that the proper understanding of all forces involved and how they interplay is crucial for solving any inclined plane problem. In our exercise, visualizing the forces and resolving them into their components is essential to find the coefficient of static friction and subsequently understand the object's behavior on the plane.