Problem 47 - College Physics

Question

\(\bullet$$\bullet\) A \(25.0-\mathrm{kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is 0.35 . (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the accel- eration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp"?

Step-by-Step Solution

Answer

(a) The box starts to slip at \( \alpha \approx 19.29^\circ \). (b) Acceleration is \( 1.33 \, \text{m/s}^2 \). (c) Final speed after sliding 5 m is \( 3.65 \, \text{m/s} \).

Step 1: Analyze the forces at play

When the box just starts to slip, the static friction force is at its maximum and equals the component of gravity along the ramp.The force of gravity acting down the ramp is given by \[ F_{g, parallel} = m imes g imes \sin(\alpha) \]The maximum static friction force is \[ f_{s, max} = \mu_s \times N \]where the normal force \( N \) is \[ N = m \times g \times \cos(\alpha) \] for the weight component perpendicular to the ramp.

Step 2: Determine the angle at which the box starts to slip

Set the forces equal since the static friction force will equal the gravitational force component when slipping starts:\[ \mu_s \times m \times g \times \cos(\alpha) = m \times g \times \sin(\alpha) \]Cancel out the common terms and solve for \( \alpha \):\[ \mu_s \times \cos(\alpha) = \sin(\alpha) \]Divide by \( \cos(\alpha) \):\[ \tan(\alpha) = \mu_s \]Substitute \( \mu_s = 0.35 \):\[ \alpha = \tan^{-1}(0.35) \approx 19.29^\circ \]

Step 3: Calculate the acceleration of the box

Once the box starts to move, the kinetic friction takes over. The force of gravity down the ramp minus kinetic friction gives the net force:\[ F_{net} = m \times g \times \sin(\alpha) - \mu_k \times m \times g \times \cos(\alpha) \]Using \( F = ma \):\[ a = g \times (\sin(\alpha) - \mu_k \times \cos(\alpha)) \]Substitute the known values:\[ a = 9.81 \times (\sin(19.29^\circ) - 0.25 \times \cos(19.29^\circ)) \]Calculate:\[ a \approx 1.33 \, \text{m/s}^2 \]

Step 4: Determine final speed after sliding down the ramp

Use the kinematic equation for velocity when starting from rest:\[ v^2 = u^2 + 2as \]Here, initial speed \( u = 0 \), acceleration \( a \approx 1.33 \, \text{m/s}^2 \), and distance \( s = 5.0 \, \text{m} \):\[ v^2 = 2 \times 1.33 \times 5.0 \]Calculate for \( v \):\[ v = \sqrt{2 \times 1.33 \times 5.0} \approx 3.65 \, \text{m/s} \]

Key Concepts

Static Friction

Static friction is the force that keeps an object at rest when it is subject to external forces. It acts in the opposite direction to the initiating force. For an object like a box on a ramp, static friction prevents it from sliding down until the ramp is tilted sufficiently. The maximum static friction is determined by the formula:\[ f_{s, max} = \mu_s \times N \]Here, \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force, which is the perpendicular force exerted by a surface on the object. In the scenario with the textbook box, as the ramp angle increases, the component of gravitational force parallel to the ramp also increases. When this component equals the maximum static frictional force, the box starts to slip.

Kinetic Friction

Once the box starts moving, static friction is no longer acting. Instead, kinetic friction comes into play. This is usually less than static friction, which means it's easier to keep an object moving than to start moving it. The kinetic friction force is determined by:\[ f_k = \mu_k \times N \]where \( \mu_k \) is the coefficient of kinetic friction. For our box, after it starts slipping, kinetic friction acts against the motion as it slides down the ramp. This opposing force is crucial in calculating the net force and resulting acceleration when the box is in motion. Knowing the kinetic friction also helps us predict how quickly the box will reach the bottom of the ramp.

Force Analysis

In this context, force analysis involves breaking down the forces acting on the box into components parallel and perpendicular to the ramp. The gravitational force can be split into:

• Parallel component: \( F_{g, parallel} = m \times g \times \sin(\alpha) \)
• Perpendicular component: \( F_{g, perpendicular} = m \times g \times \cos(\alpha) \)

Where \( m \) is the mass of the box, \( g \) is the acceleration due to gravity, and \( \alpha \) is the angle of the ramp. This analysis helps us understand how different forces balance each other out, and how to determine the angle at which static friction gives way to motion. Once the box moves, we consider net force and acceleration using these components and the friction forces.

Angle of Inclination

The angle of inclination \( \alpha \) is key in determining when the box will start to slip. It is the angle between the ramp and the horizontal ground. The calculation for the critical angle uses the relation:\[ \tan(\alpha) = \mu_s \]For our scenario, substituting \( \mu_s = 0.35 \), we calculate \( \alpha = \tan^{-1}(0.35) \approx 19.29^\circ \). This angle signifies the point where the component of gravitational force just matches the maximum static friction, leading the box to start slipping. Understanding this angle is crucial for predicting and controlling motion in physics problems involving inclined surfaces. Once the box begins to move, the dynamics shift to dealing with kinetic friction and acceleration.