Current Attempt in Progress The hydraulic oil in a car lift...

Current Attempt in Progress The hydraulic oil in a car lift has a density of $8.19 \times 10^2$ kg/m$^3$. The weight of the input piston is negligible. The radii of the input piston and output plunger are $5.87 \times 10^{-3}$ m and 0.114 m, respectively. What input force F is needed to support the 24300-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.40 m above that of the input plunger? (a) Number Units (b) Number Units

Transcript

00:02 Hi, in this question it is given that the density of the oil is 8 .3 in 10 to the power 2 kg per cubic meter.

00:10 The radius of the piston is given as 6 .21 detent to the 4 minus 3 meters that is the small r.

00:17 The radius of the plunger is given as 0 .171 meters that is the capital r.

00:23 Then the weight of the car is given as 25 ,100 utons.

00:27 We need to determine the force required to support the weight of the car when the bottom surface of the piston and the plunger are the same level.

00:35 Also we need to find the force required to support the weight of the car when the bottom surface of the plunger is 1 .4 meters above the plunger.

00:46 Let us go into the potter.

00:49 Here we need to determine the force required when the bottom surfaces of the piston and plunger are at the same level.

00:56 So let us use the pascal's equation, pascal's law and determine the force according to the pascal's law, the pressure remains same at all the points on the same level.

01:07 That means, the force required at the piston, divided by the cross -sectional area of the piston, must be equal to the weight of the car divided by the cross -sectional area of the plunger.

01:21 This can be rearranged and can be written as f is equal to small a divided by capital a, multiply with w now let us plug the given values and determine the 4x8 here this a is the pi r square and this is also pi r square this is pi in the small r square this is pi into capital r square now let us plug the given values that is pi into r square r is 6 .21 into 10 to the power minus 3 cold square divided by pi into 0 .171 square multiply with 25 ,100.

02:17 On calculation we obtained, the force required is equal to 33 .10 liters.

02:32 Now let us go into the part b...

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