Question

The angle of elevation of an inclined plane is 30°.

1. What is its mechanical advantage?
2. How much effort is required to lift a load of 100 kgf on this plane?
3. If the increase in the potential energy of the load is 5000 J, how much distance has it moved on the plane?

Answer 1

Given:

Angle of elevation (θ) = 30°

Load (L) = 100 kgf

Increase in Potential Energy (ΔPE) = 5000 J

To relate potential energy (Joules) to distance, we take 1 kgf  ≈ 10 N.

Thus, Load = 100 × 10 = 1000 N

(i) Mechanical Advantage (MA):

For an ideal inclined plane:

MA = `1/sin θ`

= `1/(sin 30°)`

= `1/0.5`

= 2

(ii) Effort Required (E):

MA = `"Load (L)"/"Effort (E)"`

2 = `100/E`

E = `100/2`

E = 50 kgf

(iii) Distance moved on the plane (d_E):

The increase in potential energy depends on the vertical height (h or d_L) raised:

ΔPE = Load (in Newtons) × d_L

5000 = 1000 × d_L

d_L = 5 m

Since VR = MA = 2 for an ideal plane:

VR = `("Distance moved on plane" (d_E))/("Vertical height" (d_L))`

`2 = d_E/5`

d_E = 2 × 5

d_E = 10 m