PRACTICE IT Use the worked example above to help you solve this problem. Suppose a block with mass 2.80 kg is resting on a ramp. If the coefficient of static friction between the block and the ramp is 0.400, what maximum angle can the ramp make with the horizontal before the block starts to slip down? 23.75 x Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. EXERCISE HINTS: GETTING STARTED I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. The ramp in the figure is roughed up and the experiment repeated. (a) What is the new coefficient of static friction if the maximum angle turns out to be 36.4°? (b) Find the magnitude of the maximum static friction force that acts on the block. N

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**EXAMPLE 4.12: A Block on a Ramp**

**GOAL**  
Apply the concept of static friction to an object resting on an incline.

**PROBLEM**  
Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down?

**STRATEGY**  
This is an application of Newton’s second law involving an object in equilibrium. Choose tilted coordinates, as in the figure. Use the fact that the block is just about to slip when the force of static friction takes its maximum value, \( f_s = \mu_s n \).

---

**SOLUTION**

1. **Write Newton’s laws for a static system in component form.**  
   The gravity force has two components.

   (1) \(\sum F_x = mg \sin \theta - \mu_s n = 0\)

   (2) \(\sum F_y = n - mg \cos \theta = 0\)

2. **Rearrange Equation (2) to get an expression for the normal force \( n \).**  
   \( n = mg \cos \theta \)

3. **Substitute the expression for \( n \) into Equation (1) and solve for \(\tan \theta\).**  
   \(\sum F_x = mg \sin \theta - \mu_s mg \cos \theta = 0 \rightarrow \tan \theta = \mu_s\)

4. **Apply the inverse tangent function to get the answer.**  
   \(\tan \theta = 0.350 \rightarrow \theta = \tan^{-1} (0.350) = 19.3^\circ\)

**Diagram Explanation**  
The diagram depicts a block on a ramp inclined at angle \(\theta\). The forces acting on the block include the gravitational force \( mg \), split into components \( mg \sin \theta \) (parallel to the ramp) and \( mg \cos \theta \) (perpendicular to the ramp). Frictional force \( f_s \) and normal force \( n \) are also illustrated. The axes are tilted to align with the ramp, facilitating the force component breakdown.
Transcribed Image Text:**EXAMPLE 4.12: A Block on a Ramp** **GOAL** Apply the concept of static friction to an object resting on an incline. **PROBLEM** Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down? **STRATEGY** This is an application of Newton’s second law involving an object in equilibrium. Choose tilted coordinates, as in the figure. Use the fact that the block is just about to slip when the force of static friction takes its maximum value, \( f_s = \mu_s n \). --- **SOLUTION** 1. **Write Newton’s laws for a static system in component form.** The gravity force has two components. (1) \(\sum F_x = mg \sin \theta - \mu_s n = 0\) (2) \(\sum F_y = n - mg \cos \theta = 0\) 2. **Rearrange Equation (2) to get an expression for the normal force \( n \).** \( n = mg \cos \theta \) 3. **Substitute the expression for \( n \) into Equation (1) and solve for \(\tan \theta\).** \(\sum F_x = mg \sin \theta - \mu_s mg \cos \theta = 0 \rightarrow \tan \theta = \mu_s\) 4. **Apply the inverse tangent function to get the answer.** \(\tan \theta = 0.350 \rightarrow \theta = \tan^{-1} (0.350) = 19.3^\circ\) **Diagram Explanation** The diagram depicts a block on a ramp inclined at angle \(\theta\). The forces acting on the block include the gravitational force \( mg \), split into components \( mg \sin \theta \) (parallel to the ramp) and \( mg \cos \theta \) (perpendicular to the ramp). Frictional force \( f_s \) and normal force \( n \) are also illustrated. The axes are tilted to align with the ramp, facilitating the force component breakdown.
**REMARKS**

It's interesting that the final result depends only on the coefficient of static friction. Notice also how similar Equations (1) and (2) are to the equations developed in previous problems. Recognizing such patterns is key to solving problems successfully.

**QUESTION**

A larger static friction constant would result in a: (Select all that apply.)

- [x] larger maximum angle.
- [ ] smaller component of gravitational force along the ramp at the maximum angle.
- [ ] larger component of normal force at the maximum angle.
- [ ] smaller maximum angle.
- [x] larger component of gravitational force along the ramp at the maximum angle.

**PRACTICE IT**

Use the worked example above to help you solve this problem. Suppose a block with mass 2.80 kg is resting on a ramp. If the coefficient of static friction between the block and the ramp is 0.400, what maximum angle can the ramp make with the horizontal before the block starts to slip down?

23.75° ❌   
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.

**EXERCISE**

Use the values from PRACTICE IT to help you work this exercise. The ramp in the figure is roughed up and the experiment repeated.

(a) What is the new coefficient of static friction if the maximum angle turns out to be 36.4°?

(b) Find the magnitude of the maximum static friction force that acts on the block. N

**HINTS** 

GETTING STARTED | I'M STUCK!
Transcribed Image Text:**REMARKS** It's interesting that the final result depends only on the coefficient of static friction. Notice also how similar Equations (1) and (2) are to the equations developed in previous problems. Recognizing such patterns is key to solving problems successfully. **QUESTION** A larger static friction constant would result in a: (Select all that apply.) - [x] larger maximum angle. - [ ] smaller component of gravitational force along the ramp at the maximum angle. - [ ] larger component of normal force at the maximum angle. - [ ] smaller maximum angle. - [x] larger component of gravitational force along the ramp at the maximum angle. **PRACTICE IT** Use the worked example above to help you solve this problem. Suppose a block with mass 2.80 kg is resting on a ramp. If the coefficient of static friction between the block and the ramp is 0.400, what maximum angle can the ramp make with the horizontal before the block starts to slip down? 23.75° ❌ Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. **EXERCISE** Use the values from PRACTICE IT to help you work this exercise. The ramp in the figure is roughed up and the experiment repeated. (a) What is the new coefficient of static friction if the maximum angle turns out to be 36.4°? (b) Find the magnitude of the maximum static friction force that acts on the block. N **HINTS** GETTING STARTED | I'M STUCK!
Expert Solution
Step 1

Practice:

Apply the equilibrium condition of the force along the y-axis.

Fy=0nmgcosθ=0n=mgcosθ

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