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FORCES at WORK Today is Component forces and friction.

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Presentation on theme: "FORCES at WORK Today is Component forces and friction."— Presentation transcript:

1 FORCES at WORK Today is Component forces and friction.

2 Friction results from the mutual contact of irregularities in the surfaces. (The atoms cling together at many points of contact. As sliding occurs, the atoms snap apart or are torn from one surface to the other.)

3 Static Friction the resistance force that must overcome to start an object in motion. Kinetic or Sliding Friction the resistance force between two surfaces already in motion Rolling Friction the resistance force between a surface and a rolling object. Fluid Friction the resistance force of a gas or liquid as an object passes through.

4 The direction of friction is ALWAYS in a direction opposing motion. An object sliding to the right, experiences friction toward the left. An object sliding down an inclined plane experiences friction up the inclined plane. Sliding friction is always less than starting friction.

5 Coefficient of friction, μ, μ = Friction force Normal force Tables of coefficients of friction are printed in references and textbooks.

6 Friction isn’t always bad Traction Brakes

7 If a force of 50 N is applied to a block of mass 10 kg and it does not move, what is the coefficient of friction for the block and table.

8 After his car stalls, George pushes slowly on his 2000 kg car at a constant speed, with a force of 300 N, what is the coefficient of sliding friction?

9 Force can be divided into COMPONENTS We will use our knowledge of trig. Functions to calculate these values. Fy = F sin Θ Fx = F cos Θ

10 Okay You Try… You are walking your dog Fido, and he is pulling on the chain at with a force 60 N at an angle of 40º What are the component forces of the pull?

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12 If the velocity is constant, and Fido has a weight of 70 N, what is the coefficient of friction between Fido and the ground?

13 What is the net force and acceleration of the box?

14 Fill in the blanks….

15 FORCES In EQUILIBRIUM STATICS

16 When forces are in EQUILIBRIUM, all the forces acting on a body are balanced and the body is NOT accelerating. Because the object is not accelerating we know The sum of the forces in horizontal is zero OR Σ Fx = 0 and Σ Fy = 0 and the sum of the forces in vertical is zero

17 MHS MHS A 50 N sign hangs by a single wire as shown. What is the tension in the wire? 1) Construct one of the triangles. 2) Calculate the part of the weight it supports 3) Use trig to find the missing side Tension in the wire is 38.9 N

18 Suppose the tension in each wire is found to be 50 N, what is the weight of the picture? 1. Draw the triangle 2. Solve for fy Since two Fy pulling up, the weight is 50 N

19 @ 45º Tension = 7.1 N @ 60º Tension = 5.8 N @ 15º Tension = 19.3 N A picture has a weight of 10 N, what is the tension in the wire at the following angles?

20 Consider the tow truck at the right. If the tensional force in the cable is 1000 N and if the cable makes a 60-degree angle with the horizontal, then what is the vertical component of force which contributes to lifting the car off the ground? The triangle The weight side W Sin Θ = opp/hyp Sin 60 = W/1000 N W = 866 N

21 A 50 kg crate is hanging on a 4 m long rope which sags 5.0 cm. What is the tension in the rope? What is the angle? sin Θ = opp/hyp Θ = 1.4 What is the tension? sin 1.4 = 250N/T tan Θ =.05m/2m T = 10,232 N tan Θ = opp/adj


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