Frictional Forces

Nonequilibriuium Applications of Newton’s Laws of Motion Non-equilibrium conditions occur when the object is accelerating and the forces acting on it are not balanced so the net force is not zero. Non-equilibrium: Fx = max and Fy = may Fx = max and Fy = may

Free Body Diagrams The free body diagram (FBD) is a simplified representation of an object, and the forces acting on it. It is called free because the diagram will show the object without its surroundings; i.e. the body is “free” of its environment. We will consider only the forces acting on our object of interest. The object is depicted as not connected to any other object – it is “free”. Label the forces appropriately. Do not include the forces that this body exerts on any other body. The best way to explain the free body diagram is to describe the steps required to construct one. Follow the procedure given below. (1) Isolate the body of interest. Draw a dotted circle around the object that separates our object from its surroundings. (2) Draw all external force vectors acting on that body. (3) You may indicate the body’s assumed direction of motion. This does not represent a separate force acting on the body. (4) Choose a convenient coordinate system.

Free Body Diagram N1 F T w1 y x y The force directions are as indicated in the diagram. The magnitudes should be in proportion if possible.

Tension This is the force transmitted through a “rope” from one end to the other. An ideal cord has zero mass, does not stretch, and the tension is the same throughout the cord. MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

T FBD for the mass M w Apply Newton’s 2nd Law to the mass M. Example (text problem 4.77): A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is attached to the wall. The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley. Find the tension in the rope from which the pulley hangs and the angle . w T x y FBD for the mass M Apply Newton’s 2nd Law to the mass M. MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

This statement is true only when  = 45 and Example continued: Apply Newton’s 2nd Law: FBD for the pulley: x y T F  This statement is true only when  = 45 and MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Example A supertanker of mass m = 1.50 X 108 kg is being towed by two tugboats. The tensions in the towing cables apply the forces T1 and T2 at equal angles of 30.0o with respect to the tanker’s axis. In addition, the tanker’s engines produce a forward drive force D whose magnitude is D = 7.50 X 104 N. Moreover, the water applies an opposing force R, whose magnitude is R = 4.00 X 104 N. The tanker moves forward with an acceleration that points along the tanker’s axis and has a magnitude of 2.00 X10-3 m/s2. Find the magnitudes of T1 and T2.

Fx = +T1cos30.0o + T2cos30.0o +D -R = max x component T1: +T1cos30.0o T2: +T2cos30.0o D: +D = 7.50 X104 N R: -R=-4.00 X104 N y component T1: +T1sin30.0o T2: -T2sin30.0o D: 0 R: 0 Fx = +T1cos30.0o + T2cos30.0o +D -R = max Fy = +T1sin30.0o – T2sin30.0o = 0 T=1.53 X 105 N

A flatbed is carrying a crate up a 10.0o hill. the coefficient of Example A flatbed is carrying a crate up a 10.0o hill. the coefficient of static friction between the truck bed and the crate is s = 0.350. Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the track. ax = 1.68 m/s2

Applying Newton’s Second Law Example: A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m1 = 3.00 kg and m2 = 1.00 kg. Find the tension in the cord connecting the two blocks as shown. F block 2 block 1 Assume that the rope stays taut so that both blocks have the same acceleration. MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Apply Newton’s 2nd Law to each block: FBD for block 2: FBD for block 1: T F w1 N1 x y x T w2 N2 y Apply Newton’s 2nd Law to each block: MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

These two equations contain the unknowns: a and T. Example continued: (1) These two equations contain the unknowns: a and T. (2) To solve for T, a must be eliminated. Solve for a in (2) and substitute in (1). MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Pick Your System Carefully x T w2 N2 y T F w1 N1 x y Include both objects in the system. Now when you sum the x-components of the forces the tensions cancel. In addition, since there is no friction, y-components do not contribute to the motion. MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Apparent Weight Stand on a bathroom scale. N FBD for the person: x y FBD for the person: Apply Newton’s 2nd Law: MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Apparent Weight is what the scale reads. The normal force is the force the scale exerts on you. By Newton’s 3rd Law this is also the force (magnitude only) you exert on the scale. A scale will read the normal force. is what the scale reads. When ay = 0, N = mg. The scale reads your true weight. When ay  0, N > mg or N < mg. In free fall ay = -g and N = 0. The person is weightless. MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Apparent Weight FBD for woman: N w Example (text problem 4.128): A woman of mass 51 kg is standing in an elevator. The elevator pushes up on her feet with 408 newtons of force. What is the acceleration of the elevator? FBD for woman: w N x y Apply Newton’s 2nd Law: (1) MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Apparent Weight Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2 Example continued: Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2 Unknown: ay Solving (1) for ay: The elevator could be (1) traveling upward with decreasing speed, or (2) traveling downward with increasing speed. The change in velocity is DOWNWARD. MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Free Body Diagram MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Hanging Picture MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Hanging Picture - Free Body Diagram mg T1 MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Hanging Picture Since this turned out to be a right triangle the simple trig functions are that is needed to find a solution. If the triangle was not a right triangle then the Law of Sines would have been needed. MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Atwood Machine and Variations MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Atwood’s Machine MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

A 2-Pulley Atwood Machine MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Three Body Problem MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Incline Plane Problems MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Single Incline Plane Problem MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010

Double Incline Plane Problem MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010