# force and pressure

The weight of a rider, driving scooter is assumed to be evenly distributed on both the tyres. The area of contact of each type with the ground is 10 cm2. If the pressure inside the tyres is 3 bar, find the mass of the rider (g = 10 ms−2)

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Pressure =Force/Area, so Force =Pressure x Area. Total contact area is 20 cm2. In SI units, this is 20/10000 m2 or 0.002 m2. 3 bar is (3 x 100000) Pa, or N/m2. So Force = 300000 x 0.002 N, or 600N. Mass in kg = 600/10. Since you do not know the mass of the scooter only, which also contributes to the pressure on the ground, so you cannot calculate mass of rider only.

26 May 2018

Given that ,P=3 bar=3*100000 pa or N/m^2A=2*10 cm^2 =20*10^-4 m^2P=F/A,3*100000=mg/A m=(300000*20*10^-4)/10m=60kgwhere m is the combination of weight of rider and scooter both

27 May 2018

P = 3 bar =3 X 10^5 PaA = 10 cm^2 = 10 X 10^-4 m^2Pressure is given byP = F/Athe force on a single tyre is F = PAthe force on two tyres will be 2F = 2PAthis force balances the weight of the personweight = force by two tyresmg = 2PAm = 2PA/gm = 2 X 3 X 10^5 X 10 X 10^-4 / 10m = 60 KgHence mass of rider is 60 Kg.

28 May 2018

Equations Used:P = F/A.F = W = m * g (assuming static equilibrium condition)Algebraic Manipulation:P*A / g = mTerms:Force (F) = Person's Weight distributed across both tires. (N)Area (A) = Surface Area Contact of both tires.(m2)Pressure (P) = total pressure in both tires. (Pa)Solution:(3 bar * 10^5 Pa/bar)* (0.002 m^2) / (10 m/s^-2) = mm = 60 kg.

30 May 2018

30 May 2018

Pressure can be defined as force acting upon unit area of P=F/AForce = mass times acceleration (F=Mxa)here acceleration is acceleration due to gravity 10ms-21 bar = 10^5 pascal = 10^5 newton/m2pascal is SI unit of pressure and newton is SI unit of forceNote: for correct result try to convert all the values in SI units hence area of each tyre in contact is 0.001m2 hence two tyres = 0.002m2hence P=M x a/AM=P x A/aM=3 x 10^5 x 0.002/10M=60 kgNote: here mass is the total weight of the rider and scooterIf the question would have been such that the excess pressure in the tyres are 3bar then it would have been mass of the rider

04 June 2018

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29 July 2018

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20 May 2019

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