Answer :
Answer:
Mechanical advantage of the ramp = [tex]\frac{980}{650} = 1.51[/tex]
Efficiency of the ramp = [tex]\frac{Actual\hspace{0.1cm} work\hspace{0.1cm} done}{Work}[/tex] = [tex]\frac{100\times 9.8\times 3}{650\times 5} = \frac{2940}{3250} = 0.9046 = 90.46%[/tex]
Explanation:
i) Mechanical advantage of the ramp = [tex]\frac{980}{650} = 1.51[/tex]
ii) Efficiency of the ramp = [tex]\frac{Actual\hspace{0.1cm} work\hspace{0.1cm} done}{Work}[/tex] = [tex]\frac{100\times 9.8\times 3}{650\times 5} = \frac{2940}{3250} = 0.9046 = 90.46%[/tex]
The mechanical advantage and the efficiency of the ramp are respectively;
330 N and 90%
Mechanical advantage and Efficiency
I have drawn an image of the force diagram on the crate and attached it.
Applying Newton's second law to the force diagram attached, we have;
F_fric + F + N + mg = 0
Projecting forces gives us;
In the x-direction;
F = F_fric + mg*sin α
In the y-direction;
N - mg*cos α = 0
Also;
F_fric = μmg*cos α
Thus;
F = (μmg*cos α) + (mg*sin α)
Mechanical advantage is;
ΔF = mg - F
ΔF = mg - mg((μ cos α) + (sin α))
Since the ramp is 3 m high and 5 m long, then by trigonometric ratios;
(sin α) = 3/5
cos α = 4/5
But we are given that F is 650 N.
Thus;
ΔF = (100 × 9.8) - 650
ΔF = 330 N
Efficiency of the ramp is;
η = (mgh/Fl) × 100%
η = (100 × 9.8 × 3/(650 × 5)) × 100%
η ≈ 90%
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