A man uses a crowbar of length 2.5 m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance 1 m from his hand. State the kind of lever used and calculate the effort needed.

Class II lever, 35 kgf

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Class I lever, 50 kgf

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Class III lever, 37.5 kgf

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Class I lever, 35 kgf

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Solution

The correct option is B
Class I lever, 50 kgf

Load arm = 2.5 – 1.0 = 1.5 m

Effort arm = 1m

Mechanical Advantage = $\frac{1.5}{1}$ = 1.5

Mechanical Advantage = $\frac{L o a d}{E f f o r t}$

1.5 = $\frac{75 k g f}{E}$

E = $\frac{75}{1.5}$ = 50 kgf, Class I Lever

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Similar questions

A man uses a crowbar of length 1.5 m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance 1 m from his hand. State the kind of lever and calculate the effort needed.

A man uses a crowbar of length 1.5m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance of 1m from his hand. (a) Draw a diagram of the arrangement showing the fulcrum (F), load (L) and effort (E) with their directions. (b) State the kind of lever. (c) Calculate (i) load arm, (ii) effort arm, (iii) mechanical advantage, and (iv) the effort needed.

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Q. A man uses a crowbar of length 1.5 m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance 1 m from his hand. Draw a diagram of the arrangement showing the fulcrum (F), load (L) and effort (E) with their directions, State the kind of lever. Calculate : (i) load arm, (ii) effort arm, (iii) mechanical advantage and (iv) the effort needed.

Q. A man uses a crowbar of length

1.5 m

to raise a load of

75 k g f

by putting a sharp edge below the bar at a distance

1 m

from his hand. Calculate the effort needed.

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A man uses a crowbar of length 2.5 m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance 1 m from his hand. State the kind of lever used and calculate the effort needed.

The correct option is B Class I lever, 50 kgf Load arm = 2.5 – 1.0 = 1.5 m Effort arm = 1m Mechanical Advantage = 1.51 = 1.5 Mechanical Advantage = LoadEffort 1.5 = 75kgfE E = 751.5 = 50 kgf, Class I Lever

The correct option is B
Class I lever, 50 kgf

Load arm = 2.5 – 1.0 = 1.5 m

Effort arm = 1m

Mechanical Advantage = $\frac{1.5}{1}$ = 1.5

Mechanical Advantage = $\frac{L o a d}{E f f o r t}$

1.5 = $\frac{75 k g f}{E}$

E = $\frac{75}{1.5}$ = 50 kgf, Class I Lever