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Sure! Let's go through the problem step-by-step using Pascal’s law to calculate the force required to lift the car.
### Given Data:
1. Cross-sectional area of the small piston, [tex]\( A_{\text{small}} \)[/tex] = 0.25 m²
2. Cross-sectional area of the large piston, [tex]\( A_{\text{large}} \)[/tex] = 5.0 m²
3. Mass of the car, [tex]\( m_{\text{car}} \)[/tex] = 1200 kg
4. Acceleration due to gravity, [tex]\( g \)[/tex] = 9.81 m/s²
### Step-by-Step Solution:
1. Calculate the force due to the weight of the car:
The force exerted by the car due to gravity, [tex]\( F_{\text{car}} \)[/tex], can be found using the equation:
[tex]\[ F_{\text{car}} = m_{\text{car}} \times g \][/tex]
Substituting the given values:
[tex]\[ F_{\text{car}} = 1200 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 11772 \, \text{N} \][/tex]
2. Apply Pascal's Law:
According to Pascal's law, the pressure applied on the small piston is transmitted undiminished to the large piston. This means:
[tex]\[ \text{Pressure on small piston} = \text{Pressure on large piston} \][/tex]
The pressure [tex]\( P \)[/tex] is given by force divided by area:
[tex]\[ \frac{F_{\text{small}}}{A_{\text{small}}} = \frac{F_{\text{large}}}{A_{\text{large}}} \][/tex]
Here, [tex]\( F_{\text{small}} \)[/tex] is the force exerted on the small piston and [tex]\( F_{\text{large}} \)[/tex] is the force required to lift the car, which is equal to [tex]\( F_{\text{car}} \)[/tex].
3. Rearrange the equation to find [tex]\( F_{\text{small}} \)[/tex]:
[tex]\[ F_{\text{small}} = F_{\text{car}} \times \frac{A_{\text{small}}}{A_{\text{large}}} \][/tex]
4. Substitute the known values:
[tex]\[ F_{\text{small}} = 11772 \, \text{N} \times \frac{0.25 \, \text{m}²}{5.0 \, \text{m}²} \][/tex]
[tex]\[ F_{\text{small}} = 11772 \, \text{N} \times 0.05 = 588.6 \, \text{N} \][/tex]
### Final Answer:
The force required to be exerted on the small piston to lift the car is 588.6 N, and the force exerted by the car due to its weight is 11772 N.
### Given Data:
1. Cross-sectional area of the small piston, [tex]\( A_{\text{small}} \)[/tex] = 0.25 m²
2. Cross-sectional area of the large piston, [tex]\( A_{\text{large}} \)[/tex] = 5.0 m²
3. Mass of the car, [tex]\( m_{\text{car}} \)[/tex] = 1200 kg
4. Acceleration due to gravity, [tex]\( g \)[/tex] = 9.81 m/s²
### Step-by-Step Solution:
1. Calculate the force due to the weight of the car:
The force exerted by the car due to gravity, [tex]\( F_{\text{car}} \)[/tex], can be found using the equation:
[tex]\[ F_{\text{car}} = m_{\text{car}} \times g \][/tex]
Substituting the given values:
[tex]\[ F_{\text{car}} = 1200 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 11772 \, \text{N} \][/tex]
2. Apply Pascal's Law:
According to Pascal's law, the pressure applied on the small piston is transmitted undiminished to the large piston. This means:
[tex]\[ \text{Pressure on small piston} = \text{Pressure on large piston} \][/tex]
The pressure [tex]\( P \)[/tex] is given by force divided by area:
[tex]\[ \frac{F_{\text{small}}}{A_{\text{small}}} = \frac{F_{\text{large}}}{A_{\text{large}}} \][/tex]
Here, [tex]\( F_{\text{small}} \)[/tex] is the force exerted on the small piston and [tex]\( F_{\text{large}} \)[/tex] is the force required to lift the car, which is equal to [tex]\( F_{\text{car}} \)[/tex].
3. Rearrange the equation to find [tex]\( F_{\text{small}} \)[/tex]:
[tex]\[ F_{\text{small}} = F_{\text{car}} \times \frac{A_{\text{small}}}{A_{\text{large}}} \][/tex]
4. Substitute the known values:
[tex]\[ F_{\text{small}} = 11772 \, \text{N} \times \frac{0.25 \, \text{m}²}{5.0 \, \text{m}²} \][/tex]
[tex]\[ F_{\text{small}} = 11772 \, \text{N} \times 0.05 = 588.6 \, \text{N} \][/tex]
### Final Answer:
The force required to be exerted on the small piston to lift the car is 588.6 N, and the force exerted by the car due to its weight is 11772 N.
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