1 / 6

Calculation of Gravitational Force and Coefficient of Friction

This document outlines examples of gravitational force calculations and the coefficient of friction using mass and distance. It includes two scenarios for gravitational force acting between students and particles, detailing the necessary parameters and equations to derive net forces using Newton's law of universal gravitation. Additionally, it explores the calculation of the coefficient of friction for objects sliding across surfaces at constant velocities, providing step-by-step methodology for determining acceleration due to gravity and frictional forces.

hien
Télécharger la présentation

Calculation of Gravitational Force and Coefficient of Friction

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. UGA EXAMPLE 1 Two students are separated in the classroom with a center to center distance of 2 m. The mass of A is 100 kg and the mass of B is 75 kg. Find the magnitude and direction of the net gravitational force acting on the students. 2 m 100 kg 75 kg F = G (m1m2 / r²) m1 = 100 kg m2 = 75 kg r = 2 m F = 6.67 x 10 –11Nm²/kg² (100kg * 75kg) / (2m)² = 1.25 E -7 N

  2. UGA EXAMPLE 2 Two particles are separated in space with a center to center distance of 0.478 m. The mass of A is 365 kg and the mass of B is 765 kg. Find the magnitude and direction of the net gravitational force acting on the particles. 0.478 m 365 kg 765 kg F = G (m1m2 / r²) m1 = 365 kg m2 = 765 kg r = 0.478 m F = 6.67 x 10 –11 Nm²/s² (365kg * 765kg) / (.478m)² = 8.15 E -5 N

  3. EXAMPLE In progress Calculate the acceleration of gravity on the earth if you know its mass to be 5.98 E 24 kg. 0.478 m 365 kg 765 kg a= G (m2/ r²) m1 = 365 kg m2 = 765 kg r = 0.478 m F = 6.67 x 10 –11 Nm²/s² (365kg * 765kg) / (.478m)² = 8.15 E -5 N

  4. Calculate μ when a 6 kg box requires 3 N of force to slide at constant velocity on a horizontal surface. v= constant a= 0 m/s² Fnet = 0 FF = 3 N Fg=Fw=FN = 60 N μ= FF/ FN FN = 60 N μ= 3 N / 60 N = .05 FF= 3 N Fpull= 3 N 6 kg Fg = 6 kg (10 m/s²) = 60 N

  5. Calculate μ when a 4 kg box requires 30 N of force to slide at constant velocity on a horizontal surface. v= constant a= 0 m/s² Fnet = 0 μ= Ff / FN FF = 30 N Fg=Fw=FN = 40 N FN = 40 N μ= 30 N / 40 N = .75 FF= 30 N Fpull= 30 N 4 kg Fg = 4 kg * 10 m/s² = 40 N

  6. Calculate μ when it takes 186 N to slide a 59.9 kg box across a horizontal slushy ice rink at an acceleration of 0 m/s2? a= 0 m/s² Fnet is 0 μ= Ff / FN FN = 599 N FF = 186 N FF= 186 N Fpull= 186 N 59.9 kg Fg=Fw=FN = 599 N Fg = 59.9 kg * 10 m/s² = 599 N μ= 186 N / 599 N = 0.31

More Related