A rescue service company has asked to design and make a model of "Jaws-of-Life" that they will use during the situation as illustrated above.

Physics
A rescue service company has asked to design and make a model of "Jaws-of-Life" that they will use during the situation as illustrated above.

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1350 N

Step 1: The Jaws-of-Life model uses a hydraulic system with Pascal's principle for force amplification. Pressure PP applied to the input piston is transmitted equally.

P=F1A1=F2A2P = \frac{F_1}{A_1} = \frac{F_2}{A_2}

where F1F_1 is input force, A1A_1 input area, F2F_2 output force, A2A_2 output area.

Step 2: Given input piston radius r1=0.5r_1 = 0.5 cm =0.005= 0.005 m, so A1=πr12A_1 = \pi r_1^2.

A1=π(0.005)2=π×2.5×10−5≈7.85×10−5 m2A_1 = \pi (0.005)^2 = \pi \times 2.5 \times 10^{-5} \approx 7.85 \times 10^{-5} \ m^2

Output piston radius r2=1.5r_2 = 1.5 cm =0.015= 0.015 m, so A2=πr22A_2 = \pi r_2^2.

A2=π(0.015)2=π×2.25×10−4≈7.07×10−4 m2A_2 = \pi (0.015)^2 = \pi \times 2.25 \times 10^{-4} \approx 7.07 \times 10^{-4} \ m^2

Step 3: Mechanical advantage MA=A2A1MA = \frac{A_2}{A_1}.

MA=7.07×10−47.85×10−5≈9MA = \frac{7.07 \times 10^{-4}}{7.85 \times 10^{-5}} \approx 9

Step 4: Input force from hand pump F1=150F_1 = 150 N.

F2=F1×MA=150×9=1350 NF_2 = F_1 \times MA = 150 \times 9 = 1350 \ N

The model exerts a spreading force of 1350 N on the car door.

1350 N

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The Jaws-of-Life model uses a hydraulic system with Pascal's principle for force amplification.

A rescue service company has asked to design and make a model of "Jaws-of-Life" that they will use during the situation as illustrated above.
Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI Answer
Step 1: The Jaws-of-Life model uses a hydraulic system with Pascal's principle for force amplification. Pressure P applied to the input piston is transmitted equally. P = (F_1)/(A_1) = (F_2)/(A_2) where F_1 is input force, A_1 input area, F_2 output force, A_2 output area. Step 2: Given input piston radius r_1 = 0.5 cm = 0.005 m, so A_1 = r_1^2. A_1 = (0.005)^2 = × 2.5 × 10^-5 ≈ 7.85 × 10^-5 \ m^2 Output piston radius r_2 = 1.5 cm = 0.015 m, so A_2 = r_2^2. A_2 = (0.015)^2 = × 2.25 × 10^-4 ≈ 7.07 × 10^-4 \ m^2 Step 3: Mechanical advantage MA = (A_2)/(A_1). MA = 7.07 × 10^-47.85 × 10^-5 ≈ 9 Step 4: Input force from hand pump F_1 = 150 N. F_2 = F_1 × MA = 150 × 9 = 1350 \ N The model exerts a spreading force of 1350 N on the car door. 1350 N