Question

In the figure provided, a 25.5° ramp in an airport has luggage sliding down to the owner. Suitcase A has a mass of 3.70 kg with a coefficient of friction of μk = 0.260 and another suitcase B has a mass of 9.69 kg with a coefficient of friction of μk = 0.110. If the ramp is 3.28 m long and the suitcase starts from rest, how much time will it take for suitcase A to reach its owner?

Answer 1

In the figure provided, a 25.5° ramp in an airport has luggage sliding down to the owner.  If the ramp is 3.28 m long and the suitcase starts from rest, the time it will take for suitcase A to reach its owner is 1.25 seconds

The diagrammatic representation of the information given in the question is attached in the image below.

From the information given:

• the angle of the ramp in the airport θ = 25.5°
• Let the mass at suitcase A be m₁ = 3.70 kg
• the coefficient of friction  μk₁ = 0.260
• Let the mass at suitcase B be m₂ = 9.69 kg
• the coefficient of friction  μk₂ = 0.110
• the distance of the suitcase (S) = 3.28 m

We are to find the time required for suitcase A to reach its owner.

To do this, let's compute the free body equation from the diagram.

• the free body equation at A can be expressed as:

• $\mathbf{m_1 g sin \theta + F_{AB} - \mu_k_1 = m_1a}$

• At B, the free body equation can be expressed as:
• $\mathbf{m_2 g sin \theta + F_{BA} - \mu_k_2 = m_2a}$  

Thus, both blocks tend to move together since the kinetic friction in A is larger than that of B.

So, summing up both equations, we have:

$\mathbf{m_1 g sin \theta-m_2 g sin \theta - \mu_k_1 - \mu_k_2= (m_1+m_2)a}$

Making acceleration a the subject, we have:

$\mathbf {a = \dfrac{m_1 g sin \theta-m_2 g sin \theta - \mu_k_1 - \mu_k_2}{(m_1 + m_2)}}$

$\mathbf {a = \dfrac{(3.70 \times 9.8 \times sin25.5) +(9.69 \times 9.8 \times sin 25.5 )- 0.260 - 0.110}{(m_1 + m_2)}}$

$\mathbf {a = \dfrac{(15.610) +(40.88)- 0.260 - 0.110}{(3.70+ 9.69)}}$

$\mathbf {a = \dfrac{56.12}{(13.39)}}$

a = 4.19 m/s²

Using the second equation of motion.

$\mathbf{S = ut + \dfrac{1}{2}at^2}$

Recall that the suitcase starts from rest;

∴

• u = 0 m/s

$\mathbf{S = 0(t) + \dfrac{1}{2}at^2}$

$\mathbf{S = \dfrac{1}{2}at^2}$

where;

• S = 3.28 m long
• a = 4.19 m/s²

$\mathbf{3.28 = \dfrac{1}{2}(4.19) (t)^2}$

By cross multiply;

3.28 × 2 = 4.19 t²

$\mathbf{t^2 = \dfrac{3.28 \times 2}{4.19}}$

$\mathbf{t =\mathbf{ \sqrt{\dfrac{3.28 \times 2}{4.19}}}}$

$\mathbf{t =\mathbf{ \sqrt{1.566}}}$

t = 1.25 seconds.

Therefore, we can conclude that it took suitcase A 1.25 seconds to reach its owner.

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