Consider a block of mass equal to 10kg sliding on an inclined plane of 30°, as shown in the figure below. The coefficient of kinetic friction between the block and the plane surface is c = 0.4 (a) Determine the value of the horizontal and vertical acceleration of the block. (b) If the block starts from rest in t=0s and when it is in the X=0 and Y=5m position, calculate what its horizontal and vertical position will be at the instant t=1s. (C) How long does the LM block take to reach the base of the tilted plane?

(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²

(b) x = 0.665 m and y = 4.62 m

(c) 3.61 s

Explanation:

(a) There are two ways we can solve this.  The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).

The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a.  Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.

Let's try the first method.  Sum of forces in the +y direction:

∑F = ma

N cos 30° + Nμ sin 30° − mg = maᵧ

N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°

Sum of forces in the +x direction:

∑F = ma

N sin 30° − Nμ cos 30° = maₓ

Substituting:

N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°

N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°

N cos 30° − mg = -N sin 30° tan 30°

N (cos 30° + sin 30° tan 30°) = mg

N = mg / (cos 30° + sin 30° tan 30°)

N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)

N = 86.6 N

Now, solving for the accelerations:

N sin 30° − Nμ cos 30° = maₓ

aₓ = N (sin 30° − μ cos 30°) / m

aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg

aₓ = 1.33 m/s²

N cos 30° + Nμ sin 30° − mg = maᵧ

aᵧ = N (cos 30° + μ sin 30°) / m − g

aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²

aᵧ = -0.770 m/s²

Now let's try the second method.

Sum of forces in the perpendicular direction:

∑F = ma

N − mg cos 30° = 0

N = mg cos 30°

Sum of forces in the parallel direction:

∑F = ma

mg sin 30° − Nμ = ma

mg sin 30° − mgμ cos 30° = ma

a = g (sin 30° − μ cos 30°)

a = (10 m/s²) (sin 30° − 0.4 cos 30°)

a = 1.536 m/s²

Solving for the accelerations:

aₓ = a cos 30°

aₓ = 1.33 m/s²

aᵧ = -a sin 30°

aᵧ = -0.770 m/s²

As you can see, the second method is faster and easier, but both methods will give you the same answer.

(b) In the x direction:

Given:

x₀ = 0 m

v₀ = 0 m/s

aₓ = 1.33 m/s²

t = 1 s

Find: x

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²

x = 0.665 m

In the y direction:

Given:

y₀ = 5 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

t = 1 s

Find: y

y = y₀ + v₀ t + ½ at²

y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²

y = 4.62 m

(c) In the y direction:

Given:

y₀ = 5 m

y = 0 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²

t = 3.61 s

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