Problem 121 - University Physics with Modern Physics

Question

A box with weight \(w\) is pulled at constant speed along a level floor by a force \(\vec{\boldsymbol{F}}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{\mathrm{k}}\) (a) In terms of \(\theta, \mu_{\mathrm{k}},\) and \(w\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25,\) calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) .(c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F\) , required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta . )\) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part (b).

Step-by-Step Solution

Answer

(a) \( F = \frac{\mu_k w}{\cos(\theta) + \mu_k \sin(\theta)} \); (b) Graph \( F \) vs. \( \theta \); (c) \( \theta \approx 14^\circ \) for minimum \( F \).

Step 1: Identify Forces

When a force \( \vec{\boldsymbol{F}} \) is applied at an angle \( \theta \), it can be decomposed into its horizontal component \( F \cos(\theta) \) and vertical component \( F \sin(\theta) \). The horizontal component helps to move the box, while the vertical component affects the normal force.

Step 2: Calculate the Normal Force

The normal force \( N \) is equal to the weight of the box minus the vertical component of the applied force: \( N = w - F \sin(\theta) \).

Step 3: Set Up Kinetic Friction Force Equation

The force of kinetic friction \( f_k \) is given by \( f_k = \mu_k N \), where \( N = w - F \sin(\theta) \).

Step 4: Equation of Motion and Solve for F

Since the box is moving at constant speed, the net horizontal force is zero. Thus \( F \cos(\theta) = f_k = \mu_k (w - F \sin(\theta)) \). Solve for \( F \), obtaining: \[ F = \frac{\mu_k w}{\cos(\theta) + \mu_k \sin(\theta)}. \]

Step 5: Calculate F for Various Angles

For \( w = 400 \text{ N} \) and \( \mu_k = 0.25 \), calculate \( F \) for \( \theta \) from \( 0^\circ \) to \( 90^\circ \) in \( 10^\circ \) increments using the formula: \[ F = \frac{100}{\cos(\theta) + 0.25 \sin(\theta)}. \]

Step 6: Graph the Results

Plot \( F \) against \( \theta \) using the results from Step 5 to visualize how \( F \) changes with \( \theta \).

Step 7: Determine Optimal Theta

Find \( \theta \) for minimum \( F \) by setting the derivative of \( F \) with respect to \( \theta \) to zero: \[ \frac{dF}{d\theta} = 0. \] Simplify and solve to get \( \tan(\theta) = \mu_k \), so \( \theta = \tan^{-1}(\mu_k) \). For \( \mu_k = 0.25 \), calculate: \( \theta = \tan^{-1}(0.25) \approx 14.0^\circ \).

Step 8: Confirm Minimum with Second Derivative

Check that the second derivative, \( \frac{d^2F}{d\theta^2} \), is positive for \( \theta = 14^\circ \) to confirm it is indeed a minimum.

Step 9: Compare with Graph

Verify that the angle \( \theta \approx 14^\circ \) corresponds to the minimum value of \( F \) in the graph from Step 6.

Key Concepts

Normal Force

The normal force is a crucial concept in physics problems involving contact between surfaces. It acts perpendicular to the surface and counteracts the weight of the object. When an external force is applied at an angle, such as in the given exercise with a force \( \vec{\boldsymbol{F}} \) at angle \( \theta \), the normal force not only balances the object's weight but also adjusts for any vertical components from the applied force.

For the box in the exercise, the normal force \( N \) is calculated by taking the weight \( w \) of the box and subtracting the vertical component of the applied force \( F \sin(\theta) \). The equation is:

• \( N = w - F \sin(\theta) \)

This equation accounts for how the vertical component of the pulling force reduces the normal force exerted by the floor. Understanding this helps to bridge the relationship between the forces acting vertically on an object.

Force Decomposition

When a force is applied at an angle, like the force \( \vec{\boldsymbol{F}} \) in our exercise, it is vital to break it down into components to better understand how it affects the movement. Force decomposition involves splitting the force into two perpendicular components: horizontal and vertical.

• The **horizontal component** is \( F \cos(\theta) \), which aids in moving the box along the floor. This component is crucial for overcoming friction and ensuring movement.


• The **vertical component** is \( F \sin(\theta) \), which influences the normal force as previously discussed.

Decomposing the force helps in setting up equations of motion, as it provides clarity on how the force affects different directions. It simplifies the analysis and calculation by focusing on individual components rather than the entire force vector.

Frictional Force Calculation

The frictional force plays a critical role when an object is moving on a surface. In this exercise, kinetic friction is what opposes the movement of the box. Understanding how to calculate the frictional force is essential for predicting how much force is required to keep the box moving at a constant speed.

The frictional force \( f_k \) is determined by multiplying the coefficient of kinetic friction \( \mu_k \) with the normal force \( N \):

• \( f_k = \mu_k N \)

Since we've established that \( N = w - F \sin(\theta) \), you can now calculate \( f_k \) based on the normal force's value. This calculation helps balance the force system as it ensures the horizontal component of the applied force equals the kinetic friction force for constant speed. Thus, the equation becomes:

• \( F \cos(\theta) = \mu_k (w - F \sin(\theta)) \)

Solving for \( F \), we find the formula:

• \( F = \frac{\mu_k w}{\cos(\theta) + \mu_k \sin(\theta)} \)

This formula gives a straightforward method to calculate the necessary force \( F \) in terms of the angle, weight, and coefficient of friction, ensuring the object moves consistently without accelerating or decelerating.