The diameter of the ram piston of a hydraulic jack is 120mm. The plunger has a diameter of 30mm and the stroke is 60mm. The MA of the lever is 12. Calculate:

Question

1) the force that must be applied to the lever to lift a load of 2Mg if the efficiency is 80%
2) the number of strokes needed for the lever to lift the load by 180mm
3) the number of strokes needed to lift the load by 180mm if there is a slip of 8%

Answer 1

To solve this problem, we’ll use several principles related to hydraulic systems and mechanical advantage.

Given Data:

• Diameter of ram piston, \( D_r = 120 , \text{mm} \) (which gives a radius of \( r_r = 60 , \text{mm} = 0.06 , \text{m} \))
• Diameter of plunger, \( D_p = 30 , \text{mm} \) (which gives a radius of \( r_p = 15 , \text{mm} = 0.015 , \text{m} \))
• Stroke of the plunger, \( s = 60 , \text{mm} = 0.06 , \text{m} \)
• Mechanical advantage (MA) of the lever, \( MA = 12 \)
• Load, \( W = 2 , \text{Mg} = 2000 , \text{kg} \) (which gives a weight of \( W = 2000 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 19620 , \text{N} \))
• Efficiency, \( \eta = 80% = 0.80 \)

Step 1: Calculate the force that must be applied to the lever

We can use the hydraulic lift equation:

\[ F_p = \frac{A_r}{A_p} \cdot W \]

Where:

• \( F_p \): Force on the plunger
• \( A_r = \pi r_r^2 \): Area of the ram piston
• \( A_p = \pi r_p^2 \): Area of the plunger

Calculating the areas:

\[ A_r = \pi (0.06^2) = \pi (0.0036) \approx 0.011304 , \text{m}^2 \] \[ A_p = \pi (0.015^2) = \pi (0.000225) \approx 0.000706858 , \text{m}^2 \]

Now, substituting these into the force equation:

\[ F_p = \frac{0.011304}{0.000706858} \cdot 19620 \] \[ F_p \approx 32.00 \cdot 19620 \approx 627840.00 , \text{N} \]

Since the efficiency of the system is 80%, the actual input force \( F_{\text{in}} \) required on the lever is given by:

\[ F_{\text{in}} = \frac{F_p}{MA \cdot \eta} = \frac{627840.00}{12 \cdot 0.80} = \frac{627840.00}{9.6} \approx 65400 , \text{N} \]

Step 2: Calculate the number of strokes needed to lift the load by 180mm

To calculate the number of strokes required, we can use the relationship between the displacement of the plunger and the ram piston.

The volume of fluid displaced by 1 stroke of the plunger:

\[ V = A_p \cdot s = 0.000706858 \cdot 0.06 \approx 0.00004241 , \text{m}^3 \]

Volume of fluid needed to lift the ram piston by \( 180 , \text{mm} = 0.18 , \text{m} \):

\[ V_r = A_r \cdot d = 0.011304 \cdot 0.18 \approx 0.00203652 , \text{m}^3 \]

The number of strokes \( N \) required to lift the load by \( 180 , \text{mm} \) is:

\[ N = \frac{V_r}{V} = \frac{0.00203652}{0.00004241} \approx 48 , \text{strokes} \]

Step 3: Calculate the number of strokes needed to lift the load by 180mm if there is a slip of 8%

With an 8% slip, the effective volume available after accounting for slip would be:

\[ V_{\text{effective}} = V \cdot (1 - \text{slip}) = V \cdot (1 - 0.08) = 0.00004241 \cdot 0.92 \approx 0.000039016 , \text{m}^3 \]

Now we recalculate the number of strokes needed:

\[ N_{\text{slip}} = \frac{V_r}{V_{\text{effective}}} = \frac{0.00203652}{0.000039016} \approx 52 , \text{strokes} \]

Summary of Results:

1. Force applied to the lever: 65400 N
2. Number of strokes to lift by 180mm: 48 strokes
3. Number of strokes considering 8% slip: 52 strokes