91Ó°ÊÓ

Static friction is the force that resists the initial motion of an object. In this problem, it prevents the crate from starting to move as you apply force through a rope. Its magnitude depends on the coefficient of static friction \(\mu_s\) and the normal force \(N\) acting on the crate.
The formula for static friction is \(f_s = \mu_s \times N\).
When pulling the crate, you pull at an angle (15 degrees), which affects how the normal force is calculated. This involves:

• Calculating the weight of the crate: \(W = mg = 68 \times 9.8 = 666.4\ \text{ N}\).
• Determining the normal force, which is the crate's weight minus the vertical component of your pulling force: \(N = W - F \times \text{sin}(15^\text{o})\).

To start moving the crate, the horizontal component of the pulling force must equal the static friction force: \(F \times \text{cos}(15^\text{o}) = 0.5 (W - F \times \text{sin}(15^\text{o}))\).

Once the crate starts moving, static friction is replaced by kinetic friction. This frictional force works to resist the motion of the crate across the floor and is typically lower than static friction.
Kinetic friction depends on the coefficient of kinetic friction \(\mu_k\) and the normal force \(N\).
The formula for kinetic friction is \(f_k = \mu_k \times N\).
In our scenario:

• The crate remains subjected to the same vertical forces, meaning the normal force is calculated similarly: \(N = 666.4 - 313.24 \times \text{sin}(15^\text{o})\) leads to \(N = 666.4 - 81.13 = 585.27\).
• The kinetic friction force thus becomes \(f_k = 0.35 \times 585.27 = 204.84 \text{ N}\).

The normal force \(N\) is an essential component in friction problems. It is the perpendicular force exerted by a surface to support the weight of the object resting on it.
For our problem, the weight of the crate (\text{W}) is calculated as: \(W = mg = 68 \times 9.8 = 666.4\ \text{ N}\).
The normal force is influenced by the vertical component of the pulling force. The vertical pulling force is \(F \times \text{sin}(15^\text{o})\).
Therefore, the normal force is given by: \(N = W - F \times \text{sin}(15^\text{o})\).
This step is crucial because the normal force determines the magnitudes of both static and kinetic friction, directly affecting how much force is needed to move the crate and how fast it accelerates.

To find the initial acceleration of the crate, we use Newton's second law: \( F_{\text{net}} = ma \)
The net force causing the acceleration is the horizontal pulling force minus the kinetic friction: \(F \times \text{cos}(15^\text{o}) - f_k \).
The force pulling the crate horizontally is: \(313.24 \times \text{cos}(15^\text{o})\).
We already calculated the kinetic friction force \(f_k = 204.84 \text{ N} \).
Therefore, net force equals: \( 313.24 \times \text{cos}(15^\text{o}) - 204.84 = 303.67 - 204.84 = 98.83\ \text{ N}\).
To find the acceleration: \(a = \frac{F_{\text{net}}}{m} = \frac{98.83}{68} \approx 1.453 \text{ m/s}^2 \).
This tells us how quickly the crate starts moving once the static friction is overcome.