Figure given below shows a wheelbarrow of mass $15kg$ carrying a load of $30kgf$ with its centre of gravity at A. The points B and C are the centre of the wheel and tip of the handle such that the horizontal distance $AB=20cm$ and $AC=40cm$.
Find the
A. load arm
B. effort arm
C. the mechanical advantage and,
D. the minimum effort needed for keeping the leg just off the ground.
Answer
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Hint: A lever is a rigid object or a beam that can rotate around a central axis which is known as a fulcrum. An input force is known as the effort force. The distance between the rigid body and the effort force applied is mentioned as the effort arm. This is from the fulcrum in order to rotate the beam. This great output force is known as the load force. The distance at which this load is applied from the fulcrum is defined as the load arm.
Complete step by step answer:
A. the load arm is the distance from the fulcrum of the rigid body to the point at which the force has been applied. Therefore here the load is given at the point A. Hence the load arm will be given as the distance from the A to F.
That is,
$\text{Load arm}=AF=20cm$
B. effort arm is the distance from the fulcrum to the point at which the effort force is applied. Therefore it can be written as,
$\text{Effort arm}=CF=40+20=60cm$
C. mechanical advantage is the ratio of the effort arm to the load arm. This can be written as,
$M.A=\dfrac{CF}{AF}=\dfrac{60}{20}=3$
Therefore the answer is obtained.
D. the minimum effort to keep the leg just off the ground.
This can be given as the ratio of the mass of the load to the mechanical advantage. Therefore we can write that,
$\text{Effort=}\dfrac{load}{M.A}=\dfrac{30+15}{3}=15kgf$
Therefore the answer for the question has been obtained.
Note: The mechanical advantage is obtained when we divide the distance from the fulcrum, the point at which the lever pivots, to the applied force by the distance from the fulcrum to the force of resistance. A machine which has a mechanical advantage of greater than one will enhance the input force.
Complete step by step answer:
A. the load arm is the distance from the fulcrum of the rigid body to the point at which the force has been applied. Therefore here the load is given at the point A. Hence the load arm will be given as the distance from the A to F.
That is,
$\text{Load arm}=AF=20cm$
B. effort arm is the distance from the fulcrum to the point at which the effort force is applied. Therefore it can be written as,
$\text{Effort arm}=CF=40+20=60cm$
C. mechanical advantage is the ratio of the effort arm to the load arm. This can be written as,
$M.A=\dfrac{CF}{AF}=\dfrac{60}{20}=3$
Therefore the answer is obtained.
D. the minimum effort to keep the leg just off the ground.
This can be given as the ratio of the mass of the load to the mechanical advantage. Therefore we can write that,
$\text{Effort=}\dfrac{load}{M.A}=\dfrac{30+15}{3}=15kgf$
Therefore the answer for the question has been obtained.
Note: The mechanical advantage is obtained when we divide the distance from the fulcrum, the point at which the lever pivots, to the applied force by the distance from the fulcrum to the force of resistance. A machine which has a mechanical advantage of greater than one will enhance the input force.
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