2Forces in EquilibriumEquilibrium 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.v = constantv = 0 m/s
3Forces in Equilibrium Forces are balanced. Not necessarily equal! Right-left and up-down components add out to zero, but forces may be different.
4Forces in EquilibriumMost problems focus on finding the equilibrant force.Equilibrant – force needed to exactly balance out other forces.If applied, results in zero acceleration (constant speed).
5Forces in EquilibriumThree 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.
6Forces in Equilibrium Now add the vectors to find the resultant. First resolve the 500. N force:horiz. = (500.N)(cos30º)horiz. = 433 Nvert. = (500. N)(sin30º)vert. = 250. NNext resolve the 300. N force:horiz. = (300. N)(cos -130º)horiz. = -193 Nvert. = (300. N)(sin -130º)vert. = N250. N30º500. N300. N-130º-230. N433 N-193 N
7Forces in Equilibrium Add the vectors to find the resultant: Resultant horizontal:433 N N = 240. NResultant vertical:250. N N = 20. NResultant magnitude:SQRT((240. N)2 + (20. N)2)SQRT( N2)241 NResultant angle:Arctan(20. N / 240. N)4.8º250. N500. N20. N-230. N433 N240. N300. N-193 N
8Forces in EquilibriumThe equilibrant force is exactly opposite the resultant force.Resultant angle: 4.8ºEquilibrant angle: 4.8º - 180ºEquilibrant angle: º241 N-175.2º
9Forces in EquilibriumAt 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?
10Forces in EquilibriumFirst, draw a force diagram to represent the situation:217 N3 N
11Forces 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.
12Forces in EquilibriumA 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?
13Forces in Equilibrium What forces are acting on the crate? WeightTwo 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.250kg30º
14Forces in Equilibrium What is the weight of the crate? Fw = mgFw = (250. kg)(9.81 m/s2)Fw = 2450 NThe 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.250kg1230 N
15Forces in EquilibriumKnowing 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 NBecause the angles are the same, the tension is the same in both cables (1420 N).250kg1230 N
16Forces in EquilibriumA 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 = 0.84More complicated solution:Draw a force diagram.Calculate weight, normal force, and friction.Calculate as Ff / FN
17Forces in Equilibrium FN Ff Fw As always, draw a force diagram first! Start by calculating the weight.Fw = (20.0 kg)(9.81 m/s2)Fw = 196 N40ºFwFfFN
18Forces in Equilibrium FN Ff Fw 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*cosFN = (196 N)(cos40º)FN = 150. N40ºFwFfFN150. N
19Forces in Equilibrium FN Ff Fw Calculate the component of the weight that is parallel to the inclined surface.Fpar = Fw*sinFpar = (196 N)(sin40º)Fpar = 126 NThe friction force must be the same as the parallel component of the weight (or the box would be accelerating).Ff = 126 N40ºFwFfFN150. N126 N126 N
20Forces in Equilibrium FN Ff Fw Finally, use the friction equation to solve for .Ff = FN(126 N) = (150. N) = (126 N)/(150. N) = 0.840Gives 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).40ºFwFfFN150. N126 N
21Forces in EquilibriumA 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?
22Forces 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.20ºFaFwFt200. NFa200. N20º
23Forces in EquilibriumThe horizontal component of the tension force is the same as the horizontal force exerted by the mother.tan20º = Fa / 200. NFa = (200. N)(tan20º)Fa = 72.8 N20ºFaFwFt200. N