3.1 The following data refer to a single hydraulic press
Area of plunger
Area of ram
Stroke of of plunger
Force applied to plunger
CALCULATE
3.1.1 The volume of liquid displaced by the plunger in 6 strokes
3.1 .2 The distance moved by the ram after ONE pumping strokes
3.13 The force exerted by the ram
3.1 .4 The mechanical advantage of the press

Ask by Pena Ellis. in South Africa

Mar 17,2025

Question

3.1 The following data refer to a single hydraulic press
Area of plunger
Area of ram
Stroke of of plunger
Force applied to plunger
CALCULATE
3.1.1 The volume of liquid displaced by the plunger in 6 strokes
3.1 .2 The distance moved by the ram after ONE pumping strokes
3.13 The force exerted by the ram
3.1 .4 The mechanical advantage of the press

Ask by Pena Ellis. in South Africa

Mar 17,2025

UpStudy AI · Accepted Answer

1. Calculate the volume of liquid displaced by the plunger in one stroke.  
 The volume displaced in one stroke is given by  
$$
V = A_p \times s_p = 0.09\,\mathrm{m}^2 \times 0.013\,\mathrm{m} = 0.00117\,\mathrm{m}^3.
$$

2. Compute the volume of liquid displaced in 6 strokes.  
 Multiply the volume for one stroke by 6:
$$
V_{6} = 6 \times 0.00117\,\mathrm{m}^3 \approx 0.00702\,\mathrm{m}^3.
$$

3. Determine the distance moved by the ram after one pumping stroke.  
 Since the liquid is incompressible, the volume displaced by the plunger equals the volume received by the ram. For the ram,  
$$
V = A_r \times s_r,
$$
where $$ s_r $$ is the distance moved by the ram in one stroke. Solving for $$ s_r $$:
$$
s_r = \frac{V}{A_r} = \frac{0.00117\,\mathrm{m}^3}{0.45\,\mathrm{m}^2} \approx 0.0026\,\mathrm{m}.
$$

4. Calculate the force exerted by the ram.  
 The pressure produced by the plunger is  
$$
P = \frac{F_p}{A_p} = \frac{300\,\mathrm{N}}{0.09\,\mathrm{m}^2} \approx 3333.33\,\mathrm{N/m}^2.
$$
This same pressure acts on the ram, so the force on the ram is  
$$
F_r = P \times A_r = 3333.33\,\mathrm{N/m}^2 \times 0.45\,\mathrm{m}^2 = 1500\,\mathrm{N}.
$$
Alternatively, using the ratio of the areas one can write:
$$
F_r = F_p \times \frac{A_r}{A_p} = 300\,\mathrm{N} \times \frac{0.45}{0.09} = 300\,\mathrm{N} \times 5 = 1500\,\mathrm{N}.
$$

5. Compute the mechanical advantage of the press.  
 Mechanical advantage (MA) is defined as the ratio of the output force to the input force:
$$
\text{MA} = \frac{F_r}{F_p} = \frac{1500\,\mathrm{N}}{300\,\mathrm{N}} = 5.
$$

Thus, the answers are:

- The volume of liquid displaced in 6 strokes is approximately $$0.00702\,\mathrm{m}^3$$.
- The distance moved by the ram after one stroke is approximately $$0.0026\,\mathrm{m}$$.
- The force exerted by the ram is $$1500\,\mathrm{N}$$.
- The mechanical advantage of the press is $$5$$.
- Volume displaced in 6 strokes: 0.00702 m³ - Distance moved by ram per stroke: 0.0026 m - Force exerted by ram: 1500 N - Mechanical advantage: 5

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Physics Questions