Question

A sailor pushes a 100.0 kg crate up a ramp that is 3.00 m high and 5.00 m long onto the deck of a ship. He exerts a 650.0 N force parallel to the ramp. What is the mechanical advantage of the ramp? What is the efficiency of the ramp? Your response should include all of your work and a free-body diagram

Answer 1

Answer on Question #57786, Physics / Mechanics | Relativity

A sailor pushes a $100.0\mathrm{kg}$ crate up a ramp that is $3.00\mathrm{m}$ high and $5.00\mathrm{m}$ long onto the deck of a ship. He exerts a $650.0\mathrm{N}$ force parallel to the ramp. What is the mechanical advantage of the ramp? What is the efficiency of the ramp? Your response should include all of your work and a free-body diagram.

Find: $\Delta F - ?\eta -?$

Given:

m=100 kg

h=3 m

l=5 m

F=650 N

g=9,8 N/kg

Solution:

Consider the forces, which acting on the crate.

Newton's Second Law:

$\sum_{i=1}^{n} \overrightarrow{F_i} = m \vec{a}(1)$

We believe that the body moves in straight lines and uniformly.

Because $\vec{a} = \vec{0}$ (2)

Write the vector sum of all forces:

$\sum_{i=1}^{n} \overrightarrow{F_i} = \vec{F} + \overrightarrow{F_{\text{frict}}} + \vec{N} + m\vec{g}(3)$ ,

where $\vec{\mathrm{F}}$ -traction force,

$\overrightarrow{\mathrm{F}_{\mathrm{frict}}}$ - friction force,

$\vec{\mathrm{N}}$ - reaction force,

mg-gravity

(2) and (3) in (1):

$$\vec{F} + \overrightarrow{F_{\mathrm{frct}}} + \vec{N} + m\vec{g} = \vec{0} \quad (4)$$

Find the projection of forces.

OX: $F - F_{\mathrm{frict}} - mg \sin \alpha = 0$ (5)

OY: $N - mg \cos \alpha = 0$ (6)

Friction force (with a particular approach):

$F_{\mathrm{frict}} = \mu N \quad (7)$

where $\mu$ – coefficient of friction ($\mu < 1$)

Of (6) $\Rightarrow N = mg \cos \alpha$ (8)

(8) in (7): $F_{\mathrm{frict}} = \mu mg \cos \alpha$ (9)

Of (5) $\Rightarrow F = F_{\mathrm{frict}} + mg \sin \alpha$ (10)

(9) in (10): $F = mg (\mu \cos \alpha + \sin \alpha)$ (11)

Expression ($\mu \cos \alpha + \sin \alpha$) $< 1$ (12)

Of (11) and (12) $\Rightarrow F = mg (\mu \cos \alpha + \sin \alpha) < mg$ (13)

Of (13) $\Rightarrow$ mechanical advantage of the ramp:

$\Delta F = mg - F$ (14)

Of $\Rightarrow \Delta F = 330 \, \text{N}$

Efficiency of the ramp:

$$\eta = \frac{A_{\mathrm{helpful}}}{A_{\mathrm{spent}}} \times 100\% \quad (15)$$

where $A_{\mathrm{helpful}}$ – helpful work,

$A_{\mathrm{spent}}$ – spent work

Helpful work: $A_{\mathrm{helpful}} = mgh$ (16),

Spent work: $A_{\mathrm{spent}} = Fl$ (17)

(16) and (17) in (15):

$$\eta = \frac{mgh}{Fl} \times 100\% \quad (18)$$

Of (18) $\Rightarrow \eta = 90\%$

Answer:

$\Delta F = 330 \, \text{N}$

$\eta = 90\%$

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