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Chapter 4: Problem 76

A \(2.00-\mathrm{kg}\) block is on a plane inclined at \(20.0^{\circ}\) with respect to the horizontal. The coefficient of static friction between the block and the plane is \(0.60 .\) a) How many forces are acting on the block? b) What is the normal force? c) Is this block moving? Explain.

Short Answer

Expert verified
There are 3 forces acting on the block: gravitational force, normal force, and frictional force. b) What is the normal force? The normal force is 18.51 N. c) Is this block moving? Explain. The block is not moving because the maximum static friction force (11.11 N) is greater than the parallel force component of gravity acting on the block (6.68 N).

Step by step solution

01

Identify forces acting on the block

To determine the number of forces acting on the block, we must first recognize the types of forces involved. There are three forces acting on the block: gravitational force (weight), the normal force (perpendicular to the inclined plane), and frictional force (opposing any motion). a) So, there are 3 forces acting on the block.
02

Resolve gravitational force into components

To analyze the normal force and frictional force, we need to resolve the gravitational force of the block into components parallel and perpendicular to the inclined plane: 1. Parallel component: \(F_{g \parallel} = mg \sin{(\theta)}\) 2. Perpendicular component: \(F_{g \perp} = mg \cos{(\theta)}\) Here, \(m = 2.00\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), and \(\theta = 20.0^{\circ}\). First, let's find \(F_{g\parallel}\) and \(F_{g\perp}\): 1. \(F_{g \parallel} = (2.00\,\text{kg})(9.81\,\text{m/s}^2)\sin{(20.0^{\circ})} = 6.68\,\text{N}\) 2. \(F_{g \perp} = (2.00\,\text{kg})(9.81\,\text{m/s}^2)\cos{(20.0^{\circ})} = 18.51\,\text{N}\)
03

Calculate the normal force

The normal force (\(F_N\)) acts perpendicular to the inclined plane and has the same magnitude as \(F_{g\perp}\), but in the opposite direction. Thus, b) The normal force is \(F_N = 18.51\,\text{N}\).
04

Determine the maximum static friction force

The maximum static friction force (\(F_{s\,max}\)) can be calculated using: \(F_{s\,max} = \mu_{s}F_N\) where, \(\mu_{s}=0.60\) (coefficient of static friction) and \(F_N = 18.51\,\text{N}\) (normal force). Now, let's calculate \(F_{s\,max}\): \(F_{s\,max} = (0.60)(18.51\,\text{N}) = 11.11\,\text{N}\)
05

Determine if the block is moving

To determine if the block will move, compare \(F_{s\,max}\) and \(F_{g\parallel}\). If the maximum static friction force (\(F_{s\,max}\)) is greater than or equal to the parallel gravitational force component (\(F_{g\parallel}\)), the block remains stationary. If \(F_{s\,max} < F_{g\parallel}\), the block will slide down the incline. We have calculated \(F_{g\parallel} = 6.68\,\text{N}\) and \(F_{s\,max} = 11.11\,\text{N}\). Here, \(F_{s\,max} > F_{g\parallel}\). c) So, the block is not moving, as the maximum static friction force is greater than the parallel force component of gravity acting on the block.

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