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Forces in Equilibrium.

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Presentation on theme: "Forces in Equilibrium."— Presentation transcript:

1 Forces in Equilibrium

2 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. v = constant v = 0 m/s

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 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. = N 250. N 30º 500. N 300. N -130º -230. N 433 N -193 N

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

8 Forces in Equilibrium The equilibrant force is exactly opposite the resultant force. Resultant angle: 4.8º Equilibrant angle: 4.8º - 180º Equilibrant angle: º 241 N -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 Forces in Equilibrium First, draw a force diagram to represent the situation: 217 N 3 N

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 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. 250kg 30º

14 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. 250kg 1230 N

15 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). 250kg 1230 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 Forces 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 N 40º Fw Ff FN

18 Forces 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*cos FN = (196 N)(cos40º) FN = 150. N 40º Fw Ff FN 150. N

19 Forces in Equilibrium FN Ff Fw
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 40º Fw Ff FN 150. N 126 N 126 N

20 Forces in Equilibrium FN Ff Fw
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). 40º Fw Ff FN 150. N 126 N

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 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. 20º Fa Fw Ft 200. N Fa 200. N 20º

23 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 20º Fa Fw Ft 200. N

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