a screw jack who's pitch is 2mm ,is used to raise a block of 700kg through a height of 30cm.the length of a tommy bar of the jar is 50cm.if the jack is 70% effecient calculate:velocity ratio, mechanical advantage,effort required,and workdone by the effort​

Respuesta :

1) The velocity ratio is [tex]500\pi[/tex]

2) The mechanical advantage is [tex]350 \pi[/tex]

3) The effort required is 6.2 N

4) The work done by the effort is 2922 J

Explanation:

1)

The velocity ratio of a screw jack is given by the equation:

[tex]VR=\frac{2\pi L}{p}[/tex]

where:

L is the length of the tommy bar

p is the pitch

For the screw jack in this problem, we have

L = 50 cm (length of the tommy bar)

p = 2 mm = 0.2 cm (pitch)

Therefore, the velocity ratio is

[tex]VR=\frac{2\pi (50)}{0.2}=500 \pi[/tex]

2)

Since the velocity ratio of the screw jack represents its ideal mechanical advantage, the efficiency of the machine is the ratio between the mechanical advantage and the velocity ratio:

[tex]\eta = \frac{MA}{VR}[/tex]

where

MA is the mechanical advantage

VR is the velocity ratio

In this problem, we have:

[tex]\eta = 0.70[/tex] (efficiency is 70%)

[tex]VR=500 \pi[/tex]

Therefore, the mechanical advantage here is

[tex]MA=\eta VR = (0.70)(500 \pi)=350\pi[/tex]

3)

The mechanical advantage of the screw jack can be written as

[tex]MA=\frac{Load}{Effort}[/tex]

where

Load is the force in output

Effort is the input force

Here we know that the Load is the weight of the block lifted by the screw jack:

[tex]Load = mg[/tex]

where

m = 700 kg is the mass

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

Substituting,

[tex]Load=(700)(9.8)=6860 N[/tex]

We also know that

[tex]MA=350 \pi[/tex]

Therefore, we can find the effort required:

[tex]Effort = \frac{Load}{MA}=\frac{6860}{350\pi}=6.2 N[/tex]

4)

The work done on the load is given by

[tex]W=(Effort)\cdot d_{eff}[/tex] (1)

where

Effort is the input force

[tex]d_{eff}[/tex] is the distance on the effort side

The velocity ratio can also be rewritten as

[tex]VR=\frac{d_{eff}}{d_{load}}[/tex]

Therefore we can rewrite the distance on the effort side as

[tex]d_{eff} = VR\cdot d_{load}[/tex]

So eq.(1) becomes

[tex]W=(effort)\cdot VR \cdot d_{load}[/tex]

where we have:

[tex]effort = 6.2 N\\VR = 500 \pi\\d_{load}=30 cm = 0.30 m[/tex]

Substituting,

[tex]W=(6.2)(500\pi)(0.30)=2922 J[/tex]

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