Calculating the Force Needed to Lift a 700 lb Car with a Hydraulic Jack

[designer rose]

1.A man has a hydraulic jack to lift his car. If the smaller piston has a diameter of .75 cm, and the larger piston a diameter of 6.0 cm, how much force will the man have to exert to lift one corner of the car, weighing 700 lb?

 

[Q_Goest]

How do you determine the force produced by pressure over an area?

 

[baba89]

To calculate the force needed to lift the car using the hydraulic jack, we can use the equation F1/A1 = F2/A2, where F1 is the force exerted on the smaller piston, A1 is the area of the smaller piston, F2 is the force exerted on the larger piston, and A2 is the area of the larger piston.

First, we need to convert the weight of the car from pounds to Newtons, as the unit of force used in the equation is Newtons. 1 pound is equivalent to 4.448 Newtons, so the weight of the car in Newtons is 700 lb x 4.448 N/lb = 3113.6 N.

Next, we can calculate the area of the smaller piston by using the formula A = πr^2, where r is the radius of the piston. The radius of the smaller piston is 0.75 cm/2 = 0.375 cm. Converting this to meters, we get 0.00375 m. Therefore, the area of the smaller piston is π(0.00375 m)^2 = 0.00004418 m^2.

Similarly, the area of the larger piston can be calculated as π(0.03 m)^2 = 0.0002827 m^2.

Now, we can plug these values into the equation F1/A1 = F2/A2 and solve for F1, which is the force that the man needs to exert on the smaller piston to lift the car. The equation becomes F1 = (F2 x A1)/A2. Plugging in the values, we get F1 = (3113.6 N x 0.00004418 m^2)/0.0002827 m^2 = 485.5 N.

Therefore, the man will need to exert a force of 485.5 Newtons on the smaller piston to lift the car weighing 700 lb using the hydraulic jack.