Question

The ratio between the length the arms of lever is 4.1 . How much force be applied to the longer arm so that a load of 20 N suspended at shorter arm can be balanced? ​

Answer 1

Answer:

The figure shows a wheel barrow of mass 15 kg carrying a load of 30 kgf

with its center of gravity at A. The points B and C are the centre of wheel

and tip of the handle such that the horizontal distance AB = 20 cm and

AC = 40 cm. Find: (a) the load arm, (b) the effort arm, (c) the mechanical

advantage, and (d) the minimum effort required to keep the leg just off

the ground.

Solution: (i) Load arm AF = 20 cm

(ii) Effort arm CF = 60 cm

(iii) Mechanical advantage M.A. = CF / AF = 60 / 20 = 3

(iv) Total load = 30 + 15 = 45 kgf

Effort = Load / M.A = (30 + 15) / 3 = 15 kgf

Explanation:

STATIC EQUILIBRIUM IN LEVERS

For all levers the effort and resistance (load) are actually just forces that are creating torques because they are trying to rotate the lever.  In order to move or hold a load the torque created by the effort must be large enough to balance the torque caused by the load. Remembering that torque depends on the distance that the force is applied from the pivot, the effort needed to balance the resistance must depend on the distances of the effort and resistance from the pivot. These distances are known as the effort arm and resistance arm (load arm). Increasing the effort arm reduces the size of the effort needed to balance the load torque. In fact, the ratio of the effort to the load is equal to the ratio of the effort arm to the load arm:

Answer 2

5 N force is required to balance the load.

Given - Ratio and load on shorter arm

Find - Load on larger arm

Solution - Let 4x and 1x be length of larger and shorter arm.

Length*Force (short arm) = Length*Force (longer arm)

1x*20 = 4x*Force

Force = $\frac{1x \times 20}{4x}$

Cancelling x and performing division

Force = 5 N

Hence, the shorter arm can be balanced by load of 5 N.

#SPJ2