A crate with mass 50 kg has a horizontal force of 95 N applied towards the right.
The crate does not move. What is the value of the force of static friction acting on the
crate if the coefficient of static friction is μs=0.40.

In this scenario, we are dealing with a crate subjected to an applied force and static friction. Since the crate does not move, the force of static friction must be equal and opposite to the applied force, preventing motion.

### Step 1: Analyze the forces
- The applied force is 95 N towards the right.
- The force of static friction is acting towards the left, opposing the applied force.
- The crate does not move, which means the force of static friction is equal to the applied force (since the static friction adjusts up to its maximum limit to prevent motion).

### Step 2: Calculate the maximum static friction force
The maximum static friction force can be calculated using the formula:
\[
f_{\text{static, max}} = \mu_s \cdot N
\]
where:
- \( f_{\text{static, max}} \) is the maximum static friction force,
- \( \mu_s = 0.40 \) is the coefficient of static friction,
- \( N \) is the normal force, which for a horizontal surface is equal to the weight of the object, \( N = mg \).

The weight of the crate is:
\[
N = m \cdot g = 50 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 490 \, \text{N}
\]

Now, calculate the maximum static friction force:
\[
f_{\text{static, max}} = 0.40 \cdot 490 \, \text{N} = 196 \, \text{N}
\]

### Step 3: Compare the forces
- The applied force is 95 N.
- The maximum possible static friction force is 196 N.

Since the applied force (95 N) is less than the maximum static friction force (196 N), the crate does not move. Therefore, the actual static friction force is exactly 95 N, balancing the applied force.

### Final Answer:
The force of static friction acting on the crate is **95 N**.

Would you like further clarification or more details?

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Here are five related questions for deeper understanding:
1. What would happen if the applied force exceeded the maximum static friction force?
2. How would the normal force change if the surface were inclined?
3. If the coefficient of static friction were lower, how would it affect the result?
4. How is the kinetic friction force different from static friction?
5. What other forces act on the crate if it were on a sloped surface?

**Tip**: The static friction force adjusts to match the applied force, but only up to its maximum value. Once this limit is exceeded, motion occurs.

A crate with mass 50 kg has a horizontal force of 95 N applied towards the right. The crate does not move. What is the value of the force of static friction acting on the crate if the coefficient of static friction is μs=0.40.

Find the static friction force for a crate with a mass of 50 kg when a 95 N horizontal force is applied. This physics problem covers key concepts such as static friction, normal force, and their corresponding formulas. Suitable for students in Grades 9-12.

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