Question

An object on an inclined plane does not slide provided the component of the object's weight parallel to the plane \(\left|\mathbf{W}_{\text {par }}\right|\) is less than or equal to the magnitude of the opposing frictional force \(\left|\mathbf{F}_{\mathrm{f}}\right| .\) The magnitude of the frictional force in turn, is proportional to the component of the object's weight perpendicular to the plane \(\left|\mathbf{W}_{\text {perp }}\right|\) (see figure). The constant of proportionality is the coefficient of static friction \(\mu>0\) a. Suppose a 100 -lb block rests on a plane that is tilted at an angle of \(\theta=20^{\circ}\) to the horizontal. Find \(\left|\mathbf{W}_{\text {par }}\right|\) and \(\left|\mathbf{W}_{\text {perp }}\right|\) b. The condition for the block not sliding is \(\left|\mathbf{W}_{\mathrm{par}}\right| \leq \mu\left|\mathbf{W}_{\mathrm{perp}}\right|\) If \(\mu=0.65,\) does the block slide? c. What is the critical angle above which the block slides with \(\mu=0.65 ?\)

Short answer

The magnitudes of the weight components parallel and perpendicular to the plane are approximately 34.2 lb and 94.0 lb, respectively. b) Will the block slide given the coefficient of static friction is 0.65? No, the block will not slide since the parallel weight component (34.2 lb) is less than the maximum frictional force (61.1 lb). c) What is the critical angle above which the block will slide? The critical angle above which the block will slide is approximately 56.3 degrees.

Step-by-step solution

Find the weight components of the block

We have the weight of the block \(|\mathbf W|=100 \text{ lb}\), and the angle of inclination \(\theta = 20^\circ\). We can now find the magnitudes of the weight components parallel and perpendicular to the plane surface using trigonometry: $$|\mathbf W_{\text{par}}| = W\sin\theta$$ $$|\mathbf W_{\text{perp}}| = W\cos\theta$$ Substitute the given values: $$|\mathbf W_{\text{par}}| = (100)\sin(20^\circ) \approx 34.2 \text{ lb}$$ $$|\mathbf W_{\text{perp}}| = (100)\cos(20^\circ) \approx 94.0 \text{ lb}$$

Determine if the block will slide or not

We are given the coefficient of static friction \(\mu = 0.65\). We can calculate the maximum frictional force \(|\mathbf F_f|\) using the formula: $$|\mathbf F_f| = \mu|\mathbf W_{\text{perp}}|$$ Now substitute the values: $$|\mathbf F_f| = (0.65)(94.0) \approx 61.1 \text{ lb}$$ The condition for the block to not slide: $$|\mathbf W_{\text{par}}| \le |\mathbf F_f|$$ Compare the values: $$34.2 \text{ lb} \le 61.1 \text{ lb}$$ Since the condition holds, the block will not slide.

Find the critical angle for sliding

We need to find the critical angle \(\theta_c\) above which the block will slide. At the critical angle, the parallel weight component equals the frictional force: $$|\mathbf W_{\text{par}}| = |\mathbf F_f|$$ $$W\sin\theta_c = \mu W\cos\theta_c$$ Divide both sides by \(W\) and rearrange: $$\tan\theta_c = \frac{1}{\mu}$$ Now, substitute the value of \(\mu\): $$\tan\theta_c = \frac{1}{0.65}$$ $$\theta_c = \arctan \left(\frac{1}{0.65}\right) \approx 56.3^\circ$$ The critical angle above which the block will slide is approximately \(56.3^\circ\).

Key concepts

Static Friction

Static friction is a type of friction that occurs between two surfaces at rest relative to each other. It's what keeps an object stationary on an inclined plane, and it resolves the forces trying to move it. In physics, static friction acts to counterbalance the component of the object's weight that would cause it to slide.
The maximum possible static frictional force can be calculated as \[|\mathbf{F}_f| = \mu |\mathbf{W}_{\text{perp}}|.\]

• \(\mu\) is the coefficient of static friction, specific to the surface materials in contact.
• \(|\mathbf{W}_{\text{perp}}|\) represents the component of the object's weight perpendicular to the surface of the plane.

In the exercise, we determined the perpendicular weight component and used it to find the maximum static frictional force. As long as the parallel weight component is less than or equal to this value, static friction will successfully prevent the block from sliding. This principle can be applied further to various static friction scenarios in physics.

Trigonometry in Physics

Trigonometry is a branch of mathematics that deals with the relationships between angles and sides of triangles. In physics, it is essential for solving problems involving inclined planes, like finding components of forces or weights.
When an object sits on an inclined plane, its weight acts straight down. However, we often need to analyze this weight in two components:

• \(|\mathbf{W}_{\text{par}}| = W\sin\theta\) represents the component of weight parallel to the plane, responsible for the tendency to slide.
• \(|\mathbf{W}_{\text{perp}}| = W\cos\theta\) is the component of weight perpendicular to the plane, influencing normal force and static friction.

By breaking down the weight using sine and cosine functions, we can calculate how much of the weight contributes to the sliding effect and how much contributes to the force counteracting this motion. These calculations are crucial for understanding the behavior of objects on inclined surfaces.

Critical Angle Calculation

The critical angle is the angle of inclination beyond which an object begins to slide. It is an important concept in physics when dealing with friction and inclined planes.
To find this angle, we equate the force component parallel to the plane with the maximum static frictional force:

• \[W\sin\theta_c = \mu W\cos\theta_c\]
• Divide by \(W\) and rearrange to form:\[\tan\theta_c = \frac{1}{\mu}\]

Solving for \(\theta_c\) involves using the arctangent function:\[\theta_c = \arctan\left(\frac{1}{0.65}\right) \approx 56.3^\circ\]This angle marks the threshold between static stability and motion. Any angle greater than this will cause the block to slide due to insufficient static friction to counterbalance the parallel component of the weight.