Problem 75
Question
A block that has a mass of \(2.0 \mathrm{~kg}\) and is \(10 \mathrm{~cm}\) wide on each side just begins to slide down an inclined plane with a \(30^{\circ}\) angle of incline ( \(\mathbf{r}\) Fig. 4.45 ). Another block of the same height and same material has base dimensions of \(20 \mathrm{~cm} \times 10 \mathrm{~cm}\) and thus a mass of \(4.0 \mathrm{~kg} .\) (a) At what critical angle will the more massive block start to slide down the plane? Why? (b) Estimate the coefficient of static friction between the block and the plane.
Step-by-Step Solution
Verified Answer
(a) The critical angle is 30°. (b) The coefficient of static friction is approximately 0.577.
1Step 1: Analyze the problem
We have two blocks with different base areas but the same height and material. The smaller block begins to slide at an angle of \(30^\circ\), which gives us the coefficient of static friction. We need to find the angle at which the larger block, with dimensions \(20 \text{ cm} \times 10 \text{ cm}\) and mass \(4.0 \text{ kg}\), begins to slide down the plane.
2Step 2: Calculate the coefficient of static friction
For the block to start sliding, the component of gravitational force parallel to the incline must equal the force of static friction. The relationship can be expressed as:\[ \mu_s = \tan(\theta) \]where \(\theta = 30^\circ\) for the smaller block. Therefore, the coefficient of static friction is:\[ \mu_s = \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \]
3Step 3: Determine the angle for the larger block to slide
The critical angle for the larger block to slide is determined by the same coefficient of static friction, as the material and surface do not change. Thus, the critical angle \( \theta_c \) for the larger block is the same as for the smaller:\[ \theta_c = \tan^{-1}(\mu_s) = 30^\circ \]
4Step 4: Estimate the coefficient of static friction
The estimated coefficient of static friction between the block and the plane is 0.577, as calculated in Step 2.
Key Concepts
Coefficient of Static FrictionInclined PlaneCritical AngleGravitational Force Component
Coefficient of Static Friction
The coefficient of static friction, often denoted as \( \mu_s \), is a measure of how much force is needed to overcome static friction and make an object start moving along a surface. This value is crucial when dealing with problems of objects on inclined planes. It represents the ratio of the maximum static frictional force that can act before motion occurs to the normal force acting on the object. Simply put, it tells us how sticky or slippery a surface is relative to another surface.
For instance, in our exercise, when the smaller block on the inclined plane begins to slide at an angle of \(30^{\circ}\), this indicates that the tangential component of gravitational force has overcome the static frictional force at that point. Using the equation \( \mu_s = \tan(\theta) \), we calculated the coefficient of static friction to be approximately 0.577. This helps predict the behavior of similar blocks under similar conditions, as only the angle at which it slides and the surface texture matter here.
Understanding \( \mu_s \) provides us with insights into how surfaces interact, which is crucial when engineering systems that rely on friction, such as brakes or conveyor belts.
For instance, in our exercise, when the smaller block on the inclined plane begins to slide at an angle of \(30^{\circ}\), this indicates that the tangential component of gravitational force has overcome the static frictional force at that point. Using the equation \( \mu_s = \tan(\theta) \), we calculated the coefficient of static friction to be approximately 0.577. This helps predict the behavior of similar blocks under similar conditions, as only the angle at which it slides and the surface texture matter here.
Understanding \( \mu_s \) provides us with insights into how surfaces interact, which is crucial when engineering systems that rely on friction, such as brakes or conveyor belts.
Inclined Plane
An inclined plane is simply a flat surface tilted at an angle, other than horizontal. This simple machine makes it easier to raise or lower a load, by spreading out the effort across a longer distance. When forces are applied on an inclined plane, they are broken down into two components that affect how objects interact with the surface.
The inclined plane in our example has an angle of \(30^{\circ}\). This angle plays a significant role in determining when objects placed on the plane will start to slide. As the angle of the incline increases, the component of gravitational force that's parallel to the surface increases, making it more likely for the object to overcome static friction and begin sliding.
Inclined planes are not only important in physics problems, but also in real-world applications, such as ramps for loading goods, roads on hills, or slides in parks.
The inclined plane in our example has an angle of \(30^{\circ}\). This angle plays a significant role in determining when objects placed on the plane will start to slide. As the angle of the incline increases, the component of gravitational force that's parallel to the surface increases, making it more likely for the object to overcome static friction and begin sliding.
Inclined planes are not only important in physics problems, but also in real-world applications, such as ramps for loading goods, roads on hills, or slides in parks.
Critical Angle
The critical angle in the context of inclined planes refers to the minimum angle at which an object begins to slide down the plane. This is the point where the component of gravitational force parallel to the plane’s surface becomes large enough to surpass the static friction holding the object in place.
In our problem, the critical angle for the smaller block was measured at \(30^{\circ}\). This was calculated using the static friction coefficient. Because both blocks in the exercise are made of the same material and on the same plane, the critical angle remains \(30^{\circ}\) for the larger block as well. Regardless of the mass or size of the block, as long as the surface material and coefficient of friction do not change, the critical angle required for sliding remains the same.
It is essential to determine this angle in various applications, such as in material handling or designing roads, to ensure safety and predictability of movement.
In our problem, the critical angle for the smaller block was measured at \(30^{\circ}\). This was calculated using the static friction coefficient. Because both blocks in the exercise are made of the same material and on the same plane, the critical angle remains \(30^{\circ}\) for the larger block as well. Regardless of the mass or size of the block, as long as the surface material and coefficient of friction do not change, the critical angle required for sliding remains the same.
It is essential to determine this angle in various applications, such as in material handling or designing roads, to ensure safety and predictability of movement.
Gravitational Force Component
When an object rests on an inclined plane, gravity acts on it in two components: one parallel to the plane and one perpendicular to it. The gravitational force component parallel to the plane is what pulls the object down the slope, while the perpendicular component presses it against the surface.
These components can be calculated by breaking down the gravitational force using trigonometry. For an angle \( \theta \), the parallel component can be calculated as \( mg \sin(\theta) \), and the perpendicular component as \( mg \cos(\theta) \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. These formulas are key to understanding how an object behaves on an incline.
In our exercise, as the angle increased to \(30^{\circ}\), the parallel component became large enough to overcome static friction and the block began to slide. This underscores the importance of understanding force components when dealing with any inclined systems, be it in solving physics problems or designing engineering works.
These components can be calculated by breaking down the gravitational force using trigonometry. For an angle \( \theta \), the parallel component can be calculated as \( mg \sin(\theta) \), and the perpendicular component as \( mg \cos(\theta) \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. These formulas are key to understanding how an object behaves on an incline.
In our exercise, as the angle increased to \(30^{\circ}\), the parallel component became large enough to overcome static friction and the block began to slide. This underscores the importance of understanding force components when dealing with any inclined systems, be it in solving physics problems or designing engineering works.
Other exercises in this chapter
Problem 71
A block is projected with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\) on a horizontal surface. If the coefficient of kinetic friction between the block and the
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A person has a choice while trying to push a crate across a horizontal pad of concrete: push it at a downward angle of \(30^{\circ},\) or pull it at an upward a
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In loading a fish delivery truck, a person pushes a block of ice up a \(20^{\circ}\) incline at constant speed. The push is \(150 \mathrm{~N}\) in magnitude and
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An object (mass \(3.0 \mathrm{~kg}\) ) slides upward on a vertical wall at constant velocity when a force \(F\) of \(60 \mathrm{~N}\) acts on it at an angle of
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