001 (part 1 of 2) 10.0 points Objects with masses of 107 kg...

001 (part 1 of 2) 10.0 points Objects with masses of 107 kg and 751 kg are separated by 0.446 m. A 63 kg mass is placed midway between them. 107 kg 63 kg 751 kg 0.446 m Find the magnitude of the net gravitational force exerted by the two larger masses on the 63 kg mass. The value of the universal gravitational constant is 6.672 !! 10^{-11} N !! m^2/kg^2. Answer in units of N. 002 (part 2 of 2) 10.0 points Leaving the distance between the 107 kg and the 751 kg masses fixed, at what distance from the 751 kg mass (other than infinitely remote ones) does the 63 kg mass experience a net force of zero? Answer in units of m.

Transcript

00:01 Hello students, in this question it is given that there are two masses of m1 is equal to 107 kg and m2 is equal to 751 kg which is separated by the distance d is equal to 0 .446 meter and a mass of m is equal to 63 kg is going to placed at the midway between these two masses.

00:30 So we have to calculate the net gravitational force.

00:39 So for this we have to calculate the half distance r is equal to d by 2 which is equal to 0 .446 divided by 2 which is equal to 0 .223 meter.

00:55 So f1 is equal to g into m1 m divided by r square therefore g is the gravitational universal gravitational constant which is given by g is equal to 6 .672 into 10 power minus 11 newton meter square per kilogram square.

01:18 Therefore it can be substituted as 6 .672 into 10 power minus 11 newton meter square per kilogram square into 107 kg into 63 kg divided by radius r is equal to 0 .223 meter square.

01:44 This is the whole square.

01:46 Therefore on calculating this we will be getting f1 is equal to 4 .497 into 10 power minus 7 newton meter square divided by 0 .049729 meter square.

02:09 So here also the kilogram and kilogram will get cancelled.

02:14 So here meter square meter square will get cancelled.

02:17 Therefore f1 is equal to 9 .043 into 10 power minus 6 newton.

02:26 Similarly f2 can be calculated.

02:30 Therefore f2 is equal to g m2 m divided by r square which is equal to 6 .672 into 10 power minus 11 newton meter square per kilogram square into 751 kg into 63 kg divided by 0 .223 the whole square meter square.

02:56 Therefore on calculating this we will be getting f2 is equal to 63 point here it is yeah 63 .4781 into 10 power minus 6 newton.

03:17 Now net force f net is equal to f2 minus f1...

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