Numericals of chapter -4( based on simple machines) of class- 6

 1. In a machine an effort of 10 kilogram force is applied to lift a load of hundred kilogram force. What is its mechanical advantage?

Ans. Given,

Effort=10kgf

Load=100kgf

Mechanical advantage=?

Mechanical advantage=load/effort.

Mechanical advantage=100kgf / 10kgf

Mechanical advantage=10.

2. The mechanical advantage of a machine is 5. How much load it can exert for the effort of 2.0 kilogram force?

Ans. Mechanical advantage=5

Effort=2 Kgf

Load=?

Mechanical advantage=load/effort

MA=L/E

MA×E = L

L= MA×E

Now putting the values of mechanical advantage and effort, we get

L=5×2 kgf

L=10 Kgf.

3. The mechanical advantage of a machine is 2. It is used to raise a load of 15 kilogram force. What effort is needed?

Ans. Mechanical advantage=2

Load=15 Kgf

Effort=?

Mechanical advantage=load/effort.

MA=L/E

MA × E = L

E = L / MA

Now putting the values of mechanical advantage and load we get,

E=15kgf / 2

E= 7.5 Kgf

4.A lever of length 100 cm has effort of 15 kilogram force at a distance of 40 cm from the fulcrum at one end. What load can be applied at its other end?

Ans. Effort=15kgf

Effort arm=40 cm

Load=?

Load arm=100 cm

Mechanical advantage=effort arm / load arm

MA= EA / LA

MA= 40 cm/ 100cm

MA=4/10

MA= 2/5

Mechanical advantage=L/E

MA = L / E

MA× E = L

L= MA × E

Now substituting the the values of effort and mechanical advantage we get,

L= 2/5 × 15 Kgf

L=2 × 3 Kgf

L= 6 Kgf.

In a a lever, fulcrum is at one end at a distance of 30 cm from the load and effort is at the other end at a distance of 90 cm from the load. Find:

a) the length of loaf arm,

b) the length of effort arm, and

c) the mechanical advantage of the lever.

Ans.  Given,

fulcrum is at one end at a distance of 30 cm from the load and effort is at the other end at a distance of 90 cm from the load.

Load arm=30cm

Effort arm=120cm

Mechanical advantage= effort arm /load arm.

MA= EA / LA

MA = 120 cm / 30 cm

MA = 12/3

MA=4.