EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 12, Problem 74P
To determine

ToCalculate: Whether there are any values for the coefficients of static friction.

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Answer to Problem 74P

There are values for the coefficients of static friction as below.

  μs,beamfloor=0.58

  μs,beamfloor=0.14

Explanation of Solution

Given information :

Length of the beam =20cm

Radius of the cylinder =4.0cm

Mass of the beam =5.0kg

Mass of the cylinder =8.0kg

The coefficient of static friction between beam and cylinder =0

The coefficients of static friction between the cylinder and the floor, and between the beam and the floor, are not zero.

Formula used :

Weight of a body can be obtained by:

  F=mg

Where, m is the mass and g is the acceleration due to gravity.

Friction force is:

  fs=μsFn

Where, μs is the coefficient of static friction and Fn is the normal force.

Calculation:

The forces that act on the beam are its weight =mg

The force of the cylinder acting along the radius of the cylinder =Fc

The normal force of the ground =Fn

Friction force fs=μsFn

The forces acting on the cylinder are its weight =Mg

The force of the beam on the cylinder acting radially inward Fcb=Fc

The normal force of the ground on the cylinder =Fnc

The force of friction fsc=μscFnc

Choose the coordinate system as in the below given figure and apply the conditions for rotational and translational equilibrium.

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 12, Problem 74P

Express μs,beamfloor in terms of fs and Fn :

  μs,beamfloor=fsFn

Express μs,cylinderfloor in terms of fsc and Fnc :

  μs,cylinderfloor=fscFnc

Apply τ=0 about an axis through the right end of the beam.

  [(10cm)cosθ]mg(15cm)Fc=0

Solve for Fc to obtain,

  Fc=[(10cm)cosθ]mg15cm

  Fc=[10cos30](5.0kg)(9.81ms-2)15Fc=28.3N

Apply Fy=0 to the beam,

  Fn + Fc cosθ mg=0

  Fn=mgFccosθ

  Fn=(5.0kg)(9.81ms-2)(28.3N)cos30Fn=24.5N

Apply Fx=0 to the beam

  fs + Fc cos(90° θ ) = 0fs = Fccos(90° θ )

  fs=Fccos(90θ)=(28.3N)cos60fs=14.2N

  Fcb is the reaction force to Fc

  Fcb=Fc=28.3N radially inward.

Apply Fy=0 to the cylinder:

  FncFcbcosθMg=0

Solve for Fnc to obtain,

  Fnc=Fcbcosθ+Mg

  Fnc=(28.3N)cos30+(8.0kg)(9.81ms-2)Fnc=103N

Apply Fx=0 to the cylinder,

  fscFcbcos(90θ)=0

  fsc=Fcbcos(90θ)fsc=28.3N cos 60fsc=14.2N

  μs,beamfloor=14.2N24.5Nμs,beamfloor=0.58

And

  μs,cylinderfloor=14.2N103Nμs,beamfloor=0.14

Conclusion:

There are values for the coefficients of static friction. They are given as below.

  μs,beamfloor=0.58

  μs,beamfloor=0.14

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