What is the magnitude of the net gravitational force on the 20.0 kg mass? What is the direction of the net gravitational force on the 20.0 kg mass? What is the magnitude of the net gravitational force on the 5.0 kg mass? I'm completely lost on this.
Each pair of masses exerts equal and opposite gravitational forces on each other according to Newton's law of gravity. (Look it up if you have to.) Calculate all the forces acting on each mass and just add them up, remembering that forces are vectors.
No. 9.8 m/s^2 is the acceleration due to Earth gravity near the Earth's surface. I assume in this problem you are to calculate the gravitional forces between these objects, ignoring any other gravitating bodies (such as the Earth). Pretend they are in outer space and use Newton's law of universal gravity.
i'm still not getting it. for the force on the 20 kg unit i have 6.67*10^-11*(20)*(10)/(.2)^2 + 6.67*10^-11*(20)*(5)/(.1)^2 giving me a net total force of 3.04 * 10^-4, which is incorrect
forces are vectors, not scalars The force from the 10 kg mass points in +y direction while the force from the 5 kg mass points in the +x direction. Add them like vectors, not numbers.
i've never really understood vector addition all that much. do i just square both numbers and add them together, then take the square root?
how do i go about finding out which direction the net gravitational force on the 20kg object is? i know it would be somewhere in between the two but i don't know how to figure out how many degrees it is.
Given the y component and x component of a vector, the angle it makes with the x axis can be found using: [tex]\tan\theta = \mbox{y-component}/\mbox{x-component}[/tex]
What is the magnitude of the net gravitational force on the 5.0 kg mass? I used Force in y= [G (10)(5) / r^2 ] sin theta =5.96 x 10 ^-8 r = 0.223606 sin theta = 0.2/ 0.223606 = Force in x = [G (5)(20)/ (0.1)^2] = 6.67 x 10^-7 magnitude = sq root of (force in y) ^2 + (force in x) ^2 = 6.6966x 10^-7? is an angle in cos or sin?