Different types of Forces (Gravitational and Normal forces)by tarunsworld20 Tags: gravitational force, gravity, newtons 3rd law, normal force, weight 

#1
Aug611, 04:58 PM

P: 1

So to give a basic idea of where I am coming from:
F = G*m_{1}*m_{2}/r^{2} In the above equation, G is the gravitational constant of 6.67 x 10^{11} Nm^{2}/kg^{2} m1 = Mass of Earth m2 = Mass of Human (which we neglect since it's very tiny compared to the mass of earth) r = distance between human's center and earth's center. Plugging in numbers will result in a value of 9.8 N Q # 1) Note, the units are N. How do you get 9.8 m/s^{2} Q # 2) So, if you wanted to be exact, the gravitational force exerted on each person is of a completely different value, because the mass of every individual varies, right? 9.8 N is the force between me and earth. Now, according to Newton's third law, Force between two objects are equal on opposite. Q # 3) So, if earth's surface is pulling me down at approximately 9.8 N, then does this mean that, I am also pulling the earth up at 9.8 N ? If so, since the masses of earth and me are completely different, the acceleration at which earth is pulling me down should be completely different from the acceleration at which I am pulling up the earth right? Q # 4) If I was exerting the same amount of gravitational force on the earth as earth was on me, then how is Normal Force playing a role in helping me to stay on the ground? Whenever I look up stuff about Normal Force, the example I get is, think of an object on a ramp. Normal Force is the upward force exerted perpendicular to the ramp. Okay cool, but here is how my train of thought goes. Gravity is pulling the object down to the surface, and the object is exerting equal and opposite amount of force of gravity to stay in balance. This is what makes the object stay on the ramp right? What role is Normal Force is playing here though? If the surface is exerting a normal force, what is it's opposing force? Free body diagram tells me its the y component of the weight of the object. Is this true? Also, while we are still talking about the object on a ramp. Let's assume the angle of the ramp to be 30 degrees. If you look at the free body diagram, you can see that, Normal force and the Y component of the weight are balancing each other out, so the object is not moving up or down. It sits on the ramp quietly. However, it starts to slide down soon. What force is making it slide down? If you look at the equation to solve for acceleration, it would be: W_{x} = m*a (Where W = weight of the object. W_{x} is the x component of the weight. In this case, W_{x} = Sin(30)*m*g) Note: In the free body diagram, W is the hypotenuse (resultant). So, isn't the above equation saying that the, object is accelerating because of the x component of weight? Detailed explanation would be very much appreciated. Thank you :) 



#2
Aug611, 05:20 PM

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[tex]g= \frac{GM}{r^2}[/tex] so that [tex]F= mg= \frac{GMm}{r^2}[/tex] [tex]F= \frac{GMm}{r^2}[/tex] has units of Newtons, [itex]kg m/s^2[/itex]. That means that having removed the mass m, we have also removed the kg units leaving only [itex]m/s^2[/itex]. g is not "9.8 N", it is [itex]9.8 m/s^2[/itex], an acceleration (NOT m/s). 



#3
Aug711, 03:53 AM

P: 595





#4
Aug711, 05:29 AM

P: 460

Different types of Forces (Gravitational and Normal forces)
Yes, f = Gm1*m2/r^2 but also the force on you is given by f = m2*a so,,,
m2*a = Gm1*m2/r^2 as you can see now, m2 cancels and you are left with the acceleration your body experiences toward the Earth, which is INDEPENDANT of your mass in a similar fashion, if you want the acceleration of the Earth toward you... m1*a = Gm1*m2/r^2 m1 cancels. Now you are left with the acceleration of the Earth toward your body and it is INDEPENDANT of the Earths mass Note:The acceleration of the Earth is way smaller than your acceleration, that's why you don't notice it. 


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