Block on a Ramp

## **Problem**

The free-body diagram (Figure 1) illustrates a block of mass

m

{\displaystyle m}

that is stationary on a ramp (inclined plane). The angle of inclination is

α

{\displaystyle \alpha}

, and the coefficient of static friction is

μ

s

{\displaystyle {\mu}\_{s} }

.

**Part 1:** Identify the forces

F
1

,

F
2

,

F
3

{\displaystyle F\_1, F\_2, F\_3}

 on the free-body diagram.

**Part 2:** Determine the formula for calculating the largest angle in which the block will remain stationary.

**Part 3:** If the coefficient of static friction is 0.8, what is the maximum angle

α

{\displaystyle \alpha}

?

## **Solution**

**Part 1**

F
1

{\displaystyle F\_1}

 is the gravitational force

F
g

{\displaystyle F\_g}

.

F
2

{\displaystyle F\_2}

 is the normal force

F
N

{\displaystyle F\_N}

.

F
3

{\displaystyle F\_3 }

 is the static friction force

F

f
s

{\displaystyle {F}\_{fs} }

.

**Part 2**

Organize the forces in this problem in two directions: perpendicular to the ramp and parallel to the ramp. Then write the equations of motion

F

n
e
t

=
m
a

{\displaystyle {F}\_{net} = ma }

. Note that the gravitational force needs to be resolved into each component. The perpendicular component is

F

g
⊥

=
m
g
cos
⁡

(
α
)

{\displaystyle {F}\_{g\perp} = mg\cos{(\alpha)} }

. The parallel component is

F

g
∥

=
m
g
sin
⁡

(
α
)

{\displaystyle {F}\_{g\parallel} = mg\sin{(\alpha)} }

.

**Perpendicular**

F

n
e
t
⊥

=
m

a

⊥

F
N

−

F

g
⊥

=
m

a

⊥

{\displaystyle
\begin{align}
{F}\_{net\perp} = m{a}\_{\perp} \\[5 pt]
F\_N - {F}\_{g\perp} = m{a}\_{\perp}
\end{align} }

Since the block is stationary in the perpendicular direction, the acceleration in the perpendicular direction is zero:

F
N

−

F
g

cos
⁡

(
α
)

=
0

F
N

=
m
g
cos
⁡

(
α
)

.

{\displaystyle
\begin{align}
F\_N - F\_g \cos{(\alpha)} = 0 \\[5pt]
F\_N = mg \cos{(\alpha)}.
\end{align}
}

**Parallel**

F

n
e
t
∥

=
m

a

∥

F

f
s

−

F

g
∥

=
m

a

∥

{\displaystyle
\begin{align}
{F}\_{net\parallel} = m{a}\_{\parallel} \\[5 pt]
{F}\_{fs} - {F}\_{g\parallel} = m{a}\_{\parallel}
\end{align} }

Since the block is also stationary in the parallel direction, the acceleration in the parallel direction is zero:

F

f
s

−

F
g

sin
⁡

(
α
)

=

{\displaystyle {F}\_{fs} - F\_g \sin{(\alpha)} = 0. }

The maximum static friction is calculated using the formula

F

f
s

=

μ

s

F
N

{\displaystyle {F}\_{fs} = {\mu}\_{s} F\_N }

; thus,

μ

s

m
g
cos
⁡

(
α
)

=
m
g
sin
⁡

(
α
)

.

{\displaystyle {\mu}\_{s} mg \cos{(\alpha)} = mg \sin{(\alpha)}. }

Solving for the coefficient of static friction

μ

s

{\displaystyle {\mu}\_{s} }

 yields

μ

s

=

sin
⁡

(
α
)

cos
⁡

(
α
)

.

{\displaystyle {\mu}\_{s} = \frac{\sin{(\alpha)}}{\cos{(\alpha)}}. }

and since

tan
⁡
(
α
# )

sin
⁡

(
α
)

cos
⁡

(
α
)

{\displaystyle \tan(\alpha) = \frac{\sin{(\alpha)}}{\cos{(\alpha)}} }

, the equation is equivalent to

μ

s

=
tan
⁡
(
α
)
.

{\displaystyle {\mu}\_{s} = \tan(\alpha). }

Finally, solving for the angle

α

{\displaystyle \alpha}

gives

# α

tan

−
1

(

μ

s

)

.

{\displaystyle \alpha = {\tan}^{-1}{({\mu}\_{s})}. }

**Part 3**

If the coefficient of static friction is 0.8,

α

=

tan

−
1

(
0.8
)

α

=

38.66

∘

.

{\displaystyle
\begin{align}
\alpha &= {\tan}^{-1}{(0.8)} \\[5 pt]
\alpha &= {38.66}^{\circ}.
\end{align}
}