Math Problem Statement

PROBLEM 56: Find the maximum angle of inclination that an inclined plane could have without a block on it sliding if the coefficient of friction is 0.3. PROBLEM 57: A 2 N weight is swinging in a vertical circle of 1 m radius and the end of the cable will break if the tension exceeds 500 N. Which of the following most nearly gives the angular velocity of the weight when the cable breaks? PROBLEM 58: Two men race around a circular track 1200 meters long. If they run in opposite direction they meet every 3 minutes where as if they run in the same direction, they are side by side every 12 minutes. Find the velocity of the slower man.

Solution

Let’s go through each problem in detail.

Problem 56

Question: Find the maximum angle of inclination that an inclined plane could have without a block on it sliding if the coefficient of friction is 0.3.

To solve this:

  1. The angle of inclination θ\theta where an object is just about to slide can be found using: tan(θ)=μ\tan(\theta) = \mu where μ=0.3\mu = 0.3 (coefficient of friction).
  2. Thus, θ=arctan(0.3)\theta = \arctan(0.3)
  3. Calculating this yields approximately: θ16.7\theta \approx 16.7^\circ So, the answer is b. 16.7°.

Problem 57

Question: A 2 N weight is swinging in a vertical circle of 1 m radius, and the cable will break if the tension exceeds 500 N. What is the angular velocity when the cable breaks?

To solve this:

  1. The maximum tension in the cable occurs at the lowest point and is given by: T=mg+mω2rT = mg + m \omega^2 r where:
    • T=500NT = 500 \, \text{N} (breaking tension),
    • mg=2Nmg = 2 \, \text{N} (weight),
    • r=1mr = 1 \, \text{m},
    • ω\omega is the angular velocity we need to find.
  2. Rearrange to solve for ω\omega: 500=2+2ω2500 = 2 + 2 \omega^2
  3. Simplifying, 498=2ω2498 = 2 \omega^2 ω2=249\omega^2 = 249 ω=24915.8rad/sec\omega = \sqrt{249} \approx 15.8 \, \text{rad/sec} So, the closest answer is a. 15.8 rad/sec.

Problem 58

Question: Two men race around a circular track 1200 meters long. If they run in opposite directions and meet every 3 minutes, what is the velocity of the slower man if the faster runs at 200 m/min?

To solve this:

  1. Let v1=200m/minv_1 = 200 \, \text{m/min} (velocity of the faster runner), and v2v_2 be the velocity of the slower runner.
  2. Since they meet every 3 minutes, their combined speed relative to each other is: v1+v2=12003=400m/minv_1 + v_2 = \frac{1200}{3} = 400 \, \text{m/min}
  3. Solving for v2v_2: v2=400200=200m/minv_2 = 400 - 200 = 200 \, \text{m/min} So, the answer is b. 200 m/min.

Would you like a deeper explanation for any of these solutions or have any specific questions?


Here are five related questions for further exploration:

  1. How would the solution to Problem 56 change if the coefficient of friction were increased?
  2. Can you derive the formula used in Problem 57 for maximum tension in a rotating object?
  3. In Problem 58, what if the runners were moving in the same direction? How would that affect the answer?
  4. How does changing the radius in Problem 57 affect the angular velocity at which the cable would break?
  5. In Problem 56, how would reducing the mass of the block impact the maximum angle?

Tip: For inclined planes, remember that the frictional force limits the angle at which sliding begins. This force is a product of the normal force and the coefficient of friction.

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Math Problem Analysis

Mathematical Concepts

Physics
Circular motion
Trigonometry
Relative speed

Formulas

tan(theta) = mu
T = mg + m * omega^2 * r
Relative velocity formula for circular motion

Theorems

Angle of inclination based on friction
Centripetal force in circular motion

Suitable Grade Level

Grade 10-12