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Forces in Equilibrium

Forces in Equilibrium. v = constant. v = 0 m/s. Forces in Equilibrium. Equilibrium exists when all the forces acting on an object are canceled out. No net force = no acceleration = constant speed. Constant speed can mean zero speed. Forces in Equilibrium. Forces are balanced.

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Forces in Equilibrium

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  1. Forces in Equilibrium

  2. v = constant v = 0 m/s Forces in Equilibrium • Equilibrium exists when all the forces acting on an object are canceled out. • No net force = no acceleration = constant speed. • Constant speed can mean zero speed.

  3. Forces in Equilibrium • Forces are balanced. • Not necessarily equal! • Right-left and up-down components add out to zero, but forces may be different.

  4. Forces in Equilibrium • Most problems focus on finding the equilibrant force. • Equilibrant – force needed to exactly balance out other forces. • If applied, results in zero acceleration (constant speed).

  5. Forces in Equilibrium • Three dogs are pulling on a frisbee. The first dog pulls with a force of 500. N at an angle of +30º. The second dog pulls with a force of 300. N at an angle of -130º. With what force (and at what angle) must a third dog pull to balance the force of the first two dogs? • First: Draw a force diagram to guide your thinking. • Second: Solve for the resultant force of the first two dogs. • Third: Find the angle opposite the resultant force for your equilibrant force.

  6. 500. N 250. N 30º 433 N 300. N -230. N -130º -193 N Forces in Equilibrium • Now add the vectors to find the resultant. • First resolve the 500. N force: • horiz. = (500.N)(cos30º) • horiz. = 433 N • vert. = (500. N)(sin30º) • vert. = 250. N • Next resolve the 300. N force: • horiz. = (300. N)(cos -130º) • horiz. = -193 N • vert. = (300. N)(sin -130º) • vert. = -230. N

  7. 250. N 20. N 240. N 433 N -230. N -193 N Forces in Equilibrium • Add the vectors to find the resultant: • Resultant horizontal: • 433 N + -193 N = 240. N • Resultant vertical: • 250. N + -230. N = 20. N • Resultant magnitude: • SQRT((240. N)2 + (20. N)2) • SQRT(58 000 N2) • 241 N • Resultant angle: • Arctan(20. N / 240. N) • 4.8º 500. N 300. N

  8. 241 N -175.2º Forces in Equilibrium • The equilibrant force is exactly opposite the resultant force. • Resultant angle: 4.8º • Equilibrant angle: 4.8º - 180º • Equilibrant angle: -175.2º

  9. Forces in Equilibrium • At a point halfway between the Earth and the Moon, a stationary rocket feels 217 N of gravitational attraction to the Earth and 3 N of attraction to the Moon. • What is the net gravitational force acting on the rocket? • How much thrust (in N) will the rocket need to exert to remain stationary, and in what direction?

  10. 3 N 217 N Forces in Equilibrium • First, draw a force diagram to represent the situation:

  11. Forces in Equilibrium • What is the net force acting on the rocket? • 217 N – 3 N = 214 N Earth-ward. • What thrust must the rocket exert to remain stationary? • 214 N Moon-ward.

  12. Forces in Equilibrium • A 250.-kg crate is being lifted vertically at a constant speed by two cables that form an angle of 30º with the vertical. • What is the weight of the crate? • What is the tension in each cable?

  13. 30º 250kg Forces in Equilibrium • What forces are acting on the crate? • Weight • Two tension forces. • If the crate is moving upward at a constant speed, what can we say about these forces? • They must cancel out to zero, or the crate would be accelerating.

  14. 1230 N 250kg Forces in Equilibrium • What is the weight of the crate? • Fw = mg • Fw = (250. kg)(9.81 m/s2) • Fw = 2450 N • The vertical components of the two tension forces must add up to 2450 N. • Since there are two tensional forces (at equal angles), the vertical component of each force is therefore 2450 / 2, or 1230 N.

  15. 1230 N 250kg Forces in Equilibrium • Knowing the vertical component and angle of one of the tension forces, we can calculate the tension force itself. • cos30º = (1230 N)/(Ft) • Ft = (1230 N) / (cos30º) • Ft = (1230 N) / (0.866) • Ft = 1420 N • Because the angles are the same, the tension is the same in both cables (1420 N).

  16. Forces in Equilibrium • A 20.0-kg box slides down a 40º ramp at constant speed. What is the coefficient of kinetic friction between the box and the ramp? • Simple solution:  = tan40º = 0.84 • More complicated solution: • Draw a force diagram. • Calculate weight, normal force, and friction. • Calculate  as Ff / FN

  17. FN Ff Fw 40º Forces in Equilibrium • As always, draw a force diagram first! • Start by calculating the weight. • Fw = (20.0 kg)(9.81 m/s2) • Fw = 196 N

  18. FN Ff Fw 40º 40º Forces in Equilibrium • It may be helpful to rotate the coordinate plane so that the “horizontal” axis runs parallel to the incline. • Next, find the normal force. • FN = Fw*cos • FN = (196 N)(cos40º) • FN = 150. N 150. N

  19. FN 150. N Ff Fw 40º 40º 126 N Forces in Equilibrium • Calculate the component of the weight that is parallel to the inclined surface. • Fpar = Fw*sin • Fpar = (196 N)(sin40º) • Fpar = 126 N • The friction force must be the same as the parallel component of the weight (or the box would be accelerating). • Ff = 126 N 126 N

  20. FN 150. N Ff 126 N Fw 40º 40º 126 N Forces in Equilibrium • Finally, use the friction equation to solve for . • Ff = FN • (126 N) = (150. N) •  = (126 N)/(150. N) •  = 0.840 • Gives the same answer we calculated before using the simple method. • Simple method only works if friction and parallel component of weight are the same (constant speed).

  21. Forces in Equilibrium • A child weighing 200. N is sitting in a swing. The child’s mother pulls the swing back so that the swing makes an angle of 20º with the vertical. • How much force must the mother exert horizontally to hold the child in that position?

  22. Fa 20º 200. N 20º Ft Fa 200. N Fw Forces in Equilibrium • First, draw a force diagram! • We know that the tension force’s vertical component is 200. N, because it is matched by the child’s weight. • The applied force must be equal to and opposite from the tension force’s horizontal component. • You can calculate the magnitude of the tension force if you like, but it isn’t strictly necessary.

  23. Fa 20º 200. N 20º Ft Fa 200. N Fw Forces in Equilibrium • The horizontal component of the tension force is the same as the horizontal force exerted by the mother. • tan20º = Fa / 200. N • Fa = (200. N)(tan20º) • Fa = 72.8 N

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