7.3 Solve friction problems on inclined surfaces | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Friction

Free-body diagrams, equilibrium, and the limit of static friction

A friction problem on a slope becomes easier when the coordinate axes follow the surface. The block’s weight points vertically downward, but its components act along and into the incline. Friction acts along the contact surface and opposes the actual or impending slipping. In this lesson, each block is modeled as a particle: its size and rotation are not included, and all forces are treated as acting at one point. This lets us solve the force balance directly. Since the forces are concurrent in this model, their moments about that point are zero.

What you will learn

Draw a free-body diagram for a body on an inclined surface.
Resolve weight and applied forces into components parallel and perpendicular to the incline.
Use equilibrium to find friction and normal forces.
Check whether the required friction is within the static-friction limit.
Verify force balance and understand moment balance for the particle model.

1. Isolate the block and choose axes

Start by defining the system: isolate the block from the incline and show every force acting on it. The weight is

W=mg

, directed vertically downward. At a rough contact, the incline exerts a normal force

N

perpendicular to the surface and a friction force

F

parallel to it. Include any applied push or pull in its actual direction.

Let the incline angle above horizontal be

\theta

. Choose the positive tangent axis up the slope and the positive normal axis away from the slope. Resolving the vertical weight into these rotated directions gives a downslope component and an into-surface component. This is ordinary vector resolution: the component along an axis is the force magnitude multiplied by the cosine of the angle between the force and that axis.

W=mg,\quad W_t=W\sin\theta,\quad W_n=W\cos\theta

The normal force is perpendicular to the contact surface; friction is parallel to it.
For an incline angle measured above horizontal, weight has downslope magnitude $W\sin\theta$ and into-surface magnitude $W\cos\theta$ .
Use the same positive directions for every force component and equilibrium equation.

2. Choose the friction direction and model

Friction opposes slipping, or the tendency to slip, at the contact. First decide which way the block would tend to move if friction could not hold it. If gravity tends to pull it down the incline, friction acts up the incline. If an applied force makes upward slipping impending, friction acts down the incline.

For a block at rest, static friction adjusts to the amount required for equilibrium, subject to a maximum. Do not set it equal to that maximum automatically. Use the inequality for an ordinary resting condition. At impending slip, friction is at its limiting value, so equality applies. If a calculated friction value is negative under your assumed direction, it acts in the opposite direction.

When a force is not parallel or perpendicular to the incline, resolve it into tangent and normal components too. A force directed into the surface increases the normal force; a force pulling away from it reduces the normal force. This matters because the friction limit depends on the normal force.

0\le F\le\mu_sN,\quad F=\mu_sN\text{ at impending slip}

At rest, static friction may be less than its maximum.
At impending slip, use the limiting-friction equality.
The normal force is found from the normal-direction equilibrium equation, not automatically set equal to the weight.

3. Write equilibrium and check the result

With tangent and normal axes chosen, write one force-equilibrium equation in each direction. Include the signed components of weight, normal force, friction, and every applied force. Solve symbolically where practical, then substitute the data with consistent units.

After solving, verify both component sums. Also check the friction condition: a resting solution is possible only if the required friction does not exceed the available static-friction limit. For the particle model used here, all forces act at the same point, so their moments about that point are zero. This moment statement follows from the model; it does not describe a finite block with forces acting at different locations.

\sum F_t=0,\quad\sum F_n=0,\quad\sum M_O=0

Force equilibrium requires zero net force in both chosen directions.
Check the friction limit after finding both $F$ and $N$ .
For the concurrent-force particle model, the moment sum about the particle point is zero.

Worked example

1. Friction required to hold a block at rest

A

10.0\,\mathrm{kg}

block rests on a

25.0^\circ

incline. A

30.0\,\mathrm{N}

force acts parallel to the incline and up the slope. The coefficient of static friction is

0.25

. Find the friction force and normal force, and determine whether the block can remain at rest. Use

g=9.81\,\mathrm{m/s^2}

.

Resolve the weight
Model the block as a particle. Gravity tends to pull it down the incline more strongly than the applied force pushes it up, so friction must act up the slope if rest is to be possible.
$W=(10.0)(9.81)=98.1\,\mathrm{N},\quad W_t=W\sin25.0^\circ,\quad W_n=W\cos25.0^\circ$
Apply force equilibrium
Take up the slope and away from the slope as positive. Tangent equilibrium determines the friction required; normal equilibrium determines the contact force.
$F+30.0-98.1\sin25.0^\circ=0,\quad N-98.1\cos25.0^\circ=0$
Check whether static friction can hold
The required friction is about $11.4\,\mathrm{N}$ . The maximum available static friction is about $22.2\,\mathrm{N}$ , so the required value is within the limit.
$F=11.4\,\mathrm{N},\quad N=88.9\,\mathrm{N},\quad\mu_sN=22.2\,\mathrm{N}$

Answer: The required friction is

11.4\,\mathrm{N}

up the incline, and the normal force is

88.9\,\mathrm{N}

. Since

11.4\,\mathrm{N}<22.2\,\mathrm{N}

, static friction can hold the block at rest.

Check: Tangent balance is

11.4+30.0-98.1\sin25.0^\circ\approx0\,\mathrm{N}

. Normal balance is

88.9-98.1\cos25.0^\circ\approx0\,\mathrm{N}

. The forces are concurrent in the particle model, so the net moment about that point is zero.

Worked example

2. Coefficient at impending downward slip

A block is just about to slide down a

30.0^\circ

incline. There is no other applied force. Find the coefficient of static friction at this limiting condition.

Set the friction direction and condition
Because the impending motion is down the slope, friction acts up it. At the instant before slipping, static friction is at its limiting value.
$F=\mu_sN$
Write tangent and normal equilibrium
The weight component along the surface is balanced by friction. Perpendicular to the surface, the normal force balances the into-surface component of weight.
$F-W\sin30.0^\circ=0,\quad N-W\cos30.0^\circ=0$
Eliminate the weight
Use the two equilibrium results in the limiting-friction relation. The weight cancels, leaving the coefficient as a function of the incline angle.
$\mu_s=\frac{W\sin30.0^\circ}{W\cos30.0^\circ}=\tan30.0^\circ=0.577$

Answer: The coefficient of static friction is approximately

0.577

.

Check: The equations give

F=W\sin30.0^\circ

and

N=W\cos30.0^\circ

, so tangent and normal force sums are both zero. The concurrent forces also have zero moment sum about the particle point.

Worked example

3. Horizontal push with impending upward slip

A block with weight

200\,\mathrm{N}

rests on a

20.0^\circ

incline. A horizontal force pushes it to the right, and it is just about to move up the slope. Given

\mu_s=0.300

, find the required horizontal force and the normal force.

Resolve the push and set limiting friction
The rightward push has an upslope component and an into-surface component. Since upward slipping is impending, friction acts down the slope and equals its limiting value.
$P_t=P\cos20.0^\circ,\quad P_n=P\sin20.0^\circ,\quad F=0.300N$
Write the equilibrium equations
The normal force balances the inward components of weight and push. Along the slope, the push balances the downslope weight component and friction.
$N=200\cos20.0^\circ+P\sin20.0^\circ,\quad P\cos20.0^\circ=200\sin20.0^\circ+0.300N$
Solve for the applied force and normal force
Substitute the normal-force expression into tangent equilibrium and solve for $P$ . Then substitute that value into normal equilibrium.
$P=\frac{200(\sin20.0^\circ+0.300\cos20.0^\circ)}{\cos20.0^\circ-0.300\sin20.0^\circ}=149\,\mathrm{N},\quad N=239\,\mathrm{N}$

Answer: The required horizontal force is approximately

149\,\mathrm{N}

, and the normal force is approximately

239\,\mathrm{N}

.

Check: Tangent balance is

149\cos20.0^\circ-200\sin20.0^\circ-0.300(239)\approx0\,\mathrm{N}

. Normal balance is

239-200\cos20.0^\circ-149\sin20.0^\circ\approx0\,\mathrm{N}

. The forces are concurrent in the particle model, so their net moment about that point is zero.

Common mistakes and how to avoid them

Setting static friction equal to

\mu_sN

in every problem where the block is at rest.

Correction: For ordinary rest, use the required friction and check that it is no greater than

\mu_sN

. Use equality only at impending slip.

Drawing friction in the same direction as the tendency to slide.

Correction: Friction acts opposite the actual or impending sliding direction along the contact surface.

Using

W\cos\theta

as the downslope weight component.

Correction: For an incline angle measured above horizontal, the downslope component is

W\sin\theta

and the into-surface component is

W\cos\theta

.

Assuming the normal force always equals the block’s weight.

Correction: Use normal-direction equilibrium and include the normal components of all applied forces.

Lesson summary

Isolate the block and use axes parallel and perpendicular to the incline.
Resolve weight into $W\sin\theta$ down the slope and $W\cos\theta$ into the surface.
Choose friction opposite the actual or impending slip direction.
Use force equilibrium to find the required friction and normal force.
Check $F\le\mu_sN$ for rest; at impending slip, use $F=\mu_sN$ .
For the particle model, all forces act at one point, so their moment sum about that point is zero.

Check your understanding

Question 1

A block on a rough incline is just about to slide down. Which way does friction act?

Down the incline
Up the incline
Perpendicular away from the incline
Vertically upward

Show answer and explanation

Up the incline

Friction opposes the impending downslope motion, so it acts up the contact surface.

Question 2

A block at rest has

N=100\,\mathrm{N}

and

\mu_s=0.40

. Which statement must be true?

The friction force is exactly $40\,\mathrm{N}$ .
The friction force cannot exceed $40\,\mathrm{N}$ .
The normal force equals the block’s weight.
The friction force is zero.

Show answer and explanation

The friction force cannot exceed

40\,\mathrm{N}

.

The maximum static friction is

\mu_sN=40\,\mathrm{N}

, but the actual friction can be smaller when less is required for equilibrium.

Question 3

For an incline angle

\theta

measured above horizontal, what is the magnitude of the weight component down the slope?

$W\cos\theta$
$W\sin\theta$
$W\tan\theta$
$W$

Show answer and explanation

W\sin\theta

Resolving the vertical weight along the tangent direction gives the downslope component

W\sin\theta

.

Key terms

Normal force: The contact force perpendicular to the surface.
Static friction: A contact force parallel to the surface that opposes slipping or its tendency while the contacting bodies remain at rest relative to each other.
Impending slip: The limiting condition just before sliding begins, when static friction has reached its maximum value.
Coefficient of static friction: The dimensionless value $\mu_s$ that sets the maximum static friction through $F_{\max}=\mu_sN$ .

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7.3 · Solve friction problems on inclined surfaces

University of Alberta ENGG 130: Engineering Mechanics: Statics

Free-body diagrams, equilibrium, and the limit of static friction

What you will learn

1. Isolate the block and choose axes

2. Choose the friction direction and model

3. Write equilibrium and check the result

1. Friction required to hold a block at rest

2. Coefficient at impending downward slip

3. Horizontal push with impending upward slip

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through ENGG 130

About this lesson and its review