Numericals on Rolling Friction
Example 1: Considering a static friction, a 100 kg crate push across a horizontal floor is at a constant speed. A coefficient of rolling friction per one the crate and floor unit is 0.02.(157 words) What is the force needed for pushing the truck?
Solution:
Here, μ = 0.02
Fr = μN
The normal force can be calculated as:
N = mg
N = 100 × 9.81
N = 981
Fr = 0.02 × 981
Fr = 19.62N
Example 2: The bike covers twenty meters on the down-slope at a continuous pace of twenty meters per second. The coefficient of rolling resistance of the bicycle tires and the road have 0.01 as the value. What is the magnitude of rolling friction (rostructural material) that acts on the bicycle?
Solution:
Here, μ = 0.01
The normal force is equal to the weight of the bicycle and the rider, which we will assume is 100 kg.
N = mg
N = 100 × 9.81
N = 981
Fr = 0.01 × 981
Fr = 9.81 N
Example 3: A box with 200 kg weight is moving straight out on the floor with the ground speed. The rolling resistance between the crate and the floor is equal to 0.03. Let’s see: what is the acceleration?
Solution:
N = mg = 200 × 9.81 = 1962 N
Fr = μ × N = 0.03 × 1962 = 58.86 N
Example 4: A rolling ball having a mass of 5 kg rides on the surface, under a rolling resistance coefficient of 0.02. Analyze the rolling friction force.
Solution:
Given:
Ball weight (W): 5 kg
Coefficient of rolling friction (μ): 0.02
N = W × g = 5 × 9.81 = 49.05 N
Fr = μ × N = 0.02 × 49.05 = 0.981 N
Example 5: Derive the rolling friction force for a rolling object with radius of curvature = 0.5 meters and at 50 kg load by making use of the relevant formula.
Solution:
Given:
Radius of curvature (r): 0.5 meters
Load (W): 50 kg
Fr = μ × W × r
N = W × g = 50 × 9.81 = 490.5N
Fr = μ × W × r = μ × 490.5 × 0.5 = 245.25μ
Example 6: A bicycle wheels along the road 30 meters in distance with a coefficient of the rolling resistance of 0.015. If the total weight of the bicycle and rider amounts to 150kg, what is the maximum value of rolling friction?
Solution:
N = mg = 150 × 9.81 = 1471.5 N
Fr = μ × N = 0.015 × 1471.5 = 22.0725 N
Rolling Friction
Rolling Friction is a frictional force that opposes rolling objects. Rolling friction is applicable where the body moves along its curved surfaces. For example, wheels in vehicles, ball bearings, etc. are examples of rolling friction.
In this article, we will learn about rolling friction, its definition, laws, formulas, causes, coefficient, and the factors that affect it. We will also learn how rolling friction is different from sliding friction.
Table of Content
- What is Rolling Friction?
- Laws of Rolling Friction
- Rolling Friction Formula
- Cause of Rolling Friction
- Coefficient of Rolling Friction
- Factors Affecting Rolling Friction
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