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3.3.4 Coefficient of Friction - Inclined Planes

When dealing with inclined planes in mechanics, the coefficient of friction plays a crucial role in determining whether an object will slide down the slope or remain stationary. Let's break this down step by step.

Understanding the Scenario

Imagine you have a block resting on a slope (an inclined plane). The slope makes an angle θ\theta with the horizontal. There are two main forces acting on the block:

  1. Gravitational Force (mg)(mg): Acts vertically downward.
  2. Normal Force (R)(R): Acts perpendicular to the surface of the slope, pushing the block upwards.

Resolving the Forces

The gravitational force can be resolved into two components:

  • Component parallel to the slope: mgsinθmg \sin \theta (this force tries to slide the block down the slope).
  • Component perpendicular to the slope: mgcosθmg \cos \theta (this force is balanced by the normal force RR

Frictional Force

infoNote

The frictional force FF opposes the motion of the block sliding down the slope. The maximum value of this frictional force is given by:

Fmax=μRF_{\text{max}} = \mu R

where μ\mu is the coefficient of friction, and RR is the normal force.

Normal Force

infoNote

Since the normal force is balancing the perpendicular component of the gravitational force:

R=mgcosθR = mg \cos \theta

Condition for Motion

infoNote

The block will start to slide down the plane if the parallel component of the gravitational force (mgsinθ)(mg \sin \theta) exceeds the maximum frictional force (μR)(\mu R). So, the condition for the block to just start moving is:

mgsinθ=μRmg \sin \theta = \mu R

Substituting R=mgcosθR = mg \cos \theta, we get:

mgsinθ=μmgcosθmg \sin \theta = \mu mg \cos \theta

Cancelling mgmg from both sides:

μ=tanθ\mu = \tan \theta

Summary

infoNote
  • Coefficient of Friction (μ)(\mu): This determines how much frictional force is available to oppose motion. On an inclined plane, μ=tanθ\mu = \tan \theta if the object is just about to slide.
  • Key Forces: The weight of the object is split into components parallel and perpendicular to the slope, influencing the friction and normal force.
  • Critical Angle: The angle θ\theta where the object starts to slide is directly related to μ\mu. If θ\theta increases beyond a certain point, the object will start to slide down the plane.

infoNote

Example Problem

Problem: A block of mass 1010 kg is resting on an inclined plane that makes an angle of 3030° with the horizontal. If the coefficient of friction between the block and the plane is 0.50.5,

Question : will the block slide down the plane?

Solution:

  1. Calculate the forces:
  • mg=10×9.8=98Nmg = 10 \times 9.8 = 98 \, \text{N}
  • Parallel force: mgsin30°=98×0.5=49Nmg \sin 30° = 98 \times 0.5 = 49 \, \text{N}
  • Normal force: mgcos30°=98×3284.9Nmg \cos 30° = 98 \times \frac{\sqrt{3}}{2} \approx 84.9 \, \text{N}
  1. Calculate the maximum frictional force:
  • Fmax=μR=0.5×84.942.45NF_{\text{max}} = \mu R = 0.5 \times 84.9 \approx 42.45 \, \text{N}
  1. Compare the forces:
  • The parallel force (49N49 N) is greater than the maximum frictional force (42.45N42.45 N), so the block will slide down the plane. This example shows how the balance between gravitational and frictional forces determines whether an object will slide down an inclined plane.

infoNote

Example: Force on a Slope with Friction

Problem Statement:

  • A box of mass 1010 kg is held in limiting equilibrium on a slope inclined at 3030^\circ by a force PP.

  • The coefficient of friction between the block and the plane is 0.35 0.35. Question

  • Find the value of PP if the block is:

  1. On the point of moving down the slope.
  2. On the point of moving up the slope.

Part (a): On the Point of Moving Down the Slope

  1. Forces acting:
  • Gravitational force component along the slope: 10gsin3010g \sin 30^\circ
  • Normal reaction force: R=10gcos30R = 10g \cos 30^\circ
  • Frictional force: Fmax=μR=0.35×RF_{\text{max}} = \mu R = 0.35 \times R
  • Force PP acting up the slope.
  1. Equations of equilibrium:
  • For vertical equilibrium:
R=10gcos30R = 10g \cos 30^\circ R=53gR = 5\sqrt{3}g
  • For horizontal equilibrium (resolving forces along the slope):
P+Fmax10gcos30=0P + F_{\text{max}} - 10g \cos 30^\circ = 0 P=5g0.35R=5g0.35×53gP = 5g - 0.35\\R = 5g - 0.35 \times 5\sqrt{3}g image
  1. Calculation:
P=(2073)4)g19.295NP = \left (\frac{20 - 7\sqrt{3})}{4}\right)g \approx 19.295 \, \text{N}

Answer (a)****:P:highlight[19.295N] P \approx :highlight[19.295 \, \text{N}]


Part (b): On the Point of Moving Up the Slope

  1. Forces acting:
  • Similar forces as in part (a), but friction now acts down the slope to oppose the motion.
  1. Equations of equilibrium:
  • For horizontal equilibrium (resolving forces along the slope):
P10gsin30Fmax=0P - 10g \sin 30^\circ - F_{\text{max}} = 0 P=5g+0.35R=5g+0.35×53gP = 5g + 0.35R = 5g + 0.35 \times 5\sqrt{3}g image
  1. Calculation:
P=5g+0.35×53g78.705NP = 5g + 0.35 \times 5\sqrt{3}g \approx 78.705 \, \text{N}

Answer (b): P:highlight[78.705N]P \approx :highlight[78.705 \, \text{N}]

:::


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