In solving two-dimensional equilibrium problems, it is essential to understand the forces acting on an object and how they interact. Unlike one-dimensional problems where forces act along a single axis, two-dimensional problems require the decomposition of forces into their components, particularly when angles are involved. However, it is important to note that torques are scalars and do not require decomposition.
Consider a classic example involving a ladder resting against a vertical wall. The ladder has a mass of 10 kilograms, uniformly distributed, and makes an angle of 53 degrees with the horizontal. The gravitational force acting on the ladder, denoted as \( mg \), acts at its midpoint. The ladder's length is 4 meters, with 2 meters on either side of the midpoint. The forces acting on the ladder include the gravitational force downward, the normal force from the ground upward, and the normal force from the wall acting horizontally. Additionally, static friction at the base of the ladder prevents it from sliding.
To analyze the equilibrium of the ladder, we can set up equations based on the sum of forces and torques. The sum of forces in the x-direction must equal zero, leading to the equation:
\[ N_{\text{top}} = F_{\text{friction}} \]
where \( N_{\text{top}} \) is the normal force at the top of the ladder and \( F_{\text{friction}} \) is the static friction force at the bottom. In the y-direction, the equation is:
\[ N_{\text{bottom}} = mg \]
where \( N_{\text{bottom}} \) is the normal force at the bottom of the ladder. Given that \( m = 10 \) kg and \( g = 10 \, \text{m/s}^2 \), we find:
\[ N_{\text{bottom}} = 100 \, \text{N} \]
Next, to find the normal force at the top of the ladder, we can use the torque equation. Choosing the bottom of the ladder as the pivot point simplifies the calculation, as it eliminates the torque due to the normal force at the bottom. The torque due to the gravitational force and the normal force at the top can be expressed as:
\[ \tau_{\text{top}} = N_{\text{top}} \cdot r_{\text{top}} \cdot \sin(37^\circ) \]
\[ \tau_{mg} = mg \cdot r_{mg} \cdot \sin(53^\circ) \]
Setting these equal gives:
\[ N_{\text{top}} \cdot 4 \cdot \sin(37^\circ) = mg \cdot 2 \cdot \sin(53^\circ) \]
Substituting \( mg = 100 \, \text{N} \) allows us to solve for \( N_{\text{top}} \), yielding approximately 37.5 N. This value also represents the static friction force at the bottom, as they are equal in this scenario.
To determine the minimum coefficient of static friction (\( \mu_s \)), we use the relationship:
\[ F_{\text{friction}} = \mu_s \cdot N_{\text{bottom}} \]
Substituting the known values gives:
\[ 37.5 = \mu_s \cdot 100 \]
Thus, \( \mu_s = 0.375 \), which is a valid coefficient of friction.
Finally, to find the total contact force at the bottom of the ladder, we consider both the normal force and the frictional force. Using the Pythagorean theorem, the total contact force can be calculated as:
\[ F_{\text{bottom}} = \sqrt{(N_{\text{bottom}})^2 + (F_{\text{friction}})^2} = \sqrt{100^2 + 37.5^2} \approx 107 \, \text{N} \]
The direction of this total force can be found using the arctangent function:
\[ \theta_{\text{bottom}} = \tan^{-1}\left(\frac{F_{\text{friction}}}{N_{\text{bottom}}}\right) = \tan^{-1}\left(\frac{37.5}{100}\right) \approx 69^\circ \]
This analysis illustrates the fundamental principles of static equilibrium in two dimensions, emphasizing the importance of understanding forces, torques, and their interactions in real-world applications.
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