Ch. 4 Forces

Explain the difference between mass and weight. Find the direction and magnitude of the normal force. Understand and explain tension. Calculate tension.

Everyday Forces Four types of forces Force of Gravity (Weight) Normal Force Applied Force Friction Force

The Four Fundamental Forces Electromagnetic Caused by interactions between protons and electrons Produces friction Gravitational The weakest force Strong nuclear force The strongest force Short range Weak nuclear force

The Force of Gravity (Fg) Also called the weight of an object Weight is a scalar quantity The force exerted by the Earth on an object

Two Kinds of Mass Inertial mass: the net force on an object divided by its acceleration. m = Fnet / a Gravitational mass: Compare the gravitational attraction of an unknown mass to that of a known mass, usually with a balance. If it balances, the masses are equal. m ? Einstein asserted that these two kinds of masses are equivalent. Balance

The Force of Gravity (Fg) Depends on the location Objects weigh less at higher altitudes than at sea level Objects weigh less at lower latitudes because the Earth bulges slightly at the equator

The Normal Force (FN) The force that is perpendicular to the contact surface Counteracts gravity but isn’t always the exact opposite

The Applied Force (Fa) Force applied to an object to make and keep it moving When the applied force is removed, the object stops moving

The Force of Friction (Ff) Friction opposes the applied force Two types Static Friction Kinetic Friction

The Net Force (ΣF) ΣF = Fa - Fk The difference between the applied force and the friction force ΣF = Fa - Fk

Solving Find Fg Find Fax and Fay Find FN All upward forces = all downward forces You will either add or subtract Fg and Fay Find Fs or Fk (depending on the problem) Find net force (ΣF) then find acceleration Or If moving with constant velocity where Fk = Fax, find μk

A student moves a box by attaching a rope to the box and pulling with a force of 90.0 N at an angle of 30.0°. The box has a mass of 20.0 kg, and the coefficient of friction between the box and sidewalk is 0.50. Find the acceleration of the box.

A box of books weighing 307 N moves with a constant velocity across the floor when it is pushed with a force of 364 N exerted downward at an angle of 38.1° below the horizontal. Find the μk between the box and the floor.

A student moves a box by attaching a rope to the box and pulling with a force of 90.0 N at an angle of 30.0o. The box has a mass of 20.0 kg, and the coefficient of friction between the box and sidewalk is 0.50. Find the acceleration of the box.

A student moves a box by attaching a rope to the box and pulling with a force of 90.0 N at an angle of 30.0o. The box has a mass of 20.0 kg, and the coefficient of friction between the box and sidewalk is 0.50. Find the acceleration of the box.

Tension Force When a string or rope or wire pulls on an object, it exerts a contact force called the tension force. The tension force is in the direction of the string or rope. A rope is made of atoms joined together by molecular bonds. Molecular bonds can be modeled as tiny springs holding the atoms together. Tension is a result of many molecular springs stretching ever so slightly. Slide 5-32

Box / Tension Problem 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N A force is applied to a box that is connected to other boxes by ropes. The whole system is accelerating to the left. The problem is to find the tensions in the ropes. We can apply the 2nd Law to each box individually as well as to the whole system.

Box / Tension Analysis 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N T1 pulls on the 8-kg box to the right just as hard as it pulls on the middle box to the left. T1 must be < 38 N, or the 8-kg box couldn’t accelerate. T2 pulls on the middle box to the right just as hard as it pulls on the 6-kg box to the left. T1 must be > T2 or the middle box couldn’t accelerate.

Free Body Diagram – system 19 kg 38 N mg N For convenience, we’ll choose left to be the positive direction. The total mass of all three boxes is 19 kg. N and mg cancel out. Fnet = m a implies a = 2.0 m/s2 Since the ropes don’t stretch, a will be 2.0 m/s2 for all three boxes.

Free Body Diagram – right box N and mg cancel out. For this particular box, Fnet = m a implies: T2 = 6a = 6(2) = 12 N. (Remember, a = 2 m/s2 for all three boxes.) 6 kg T2 N mg 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N

Free Body Diagram – middle box N and mg cancel out again. Fnet = m a implies: T1 – T2 = 5a. So, T1 – 12 = 5(2), and T1 = 22 N 5 kg T2 = 12 N T1 mg N 8 kg 5 kg 6 kg T1 frictionless floor T2 38 N

Free Body Diagram – left box Let’s check our work using the left box. N and mg cancel out here too. Fnet = ma implies: 38 - 22 = ma = 8(2). 16 = 16. N 38 N T1 = 22 N 8 kg mg T2 T1 38 N 5 kg 6 kg 8 kg

Example 7.5 Pulling a Rope Slide 7-55

Example 7.5 Pulling a Rope Slide 7-56

Example 7.5 Pulling a Rope Slide 7-57

The rope’s tension is the same in both situations. Example 7.5 Pulling a Rope The rope’s tension is the same in both situations. Slide 7-58

The Massless String Approximation Often in problems the mass of the string or rope is much less than the masses of the objects that it connects. In such cases, we can adopt the following massless string approximation: Slide 7-59

The Massless String Approximation Two blocks are connected by a massless string, as block B is pulled to the right. Forces and act as if they are an action/reaction pair: All a massless string does is transmit a force from A to B without changing the magnitude of that force. For problems in this book, you can assume that any strings or ropes are massless unless it explicitly states otherwise. Slide 7-60

Atwood Device m1 m2 m1g T m2g Assume m1 < m2 and that the clockwise direction is +. If the rope & pulley have negligible mass, and if the pulley is frictionless, then T is the same throughout the rope. If the rope doesn’t stretch, a is the same for both masses.

Atwood Analysis a = T m2 – m1 m1 + m2 m1g g m2g Remember, clockwise has been defined as +. m1 m2 m1g T m2g 2nd Law on m1: T - m1g = m1a 2nd Law on m2: m2g - T = m2 a Add equations: m2g – m1g = m1a + m2 a (The T ’s cancel out.) Solve for a: m2 – m1 m1 + m2 a = g

Atwood as a system T m1g m2g Treated as a system (rope & both masses), tension is internal and the T ’s cancel out (one clock-wise, one counterclockwise). Fnet = (total mass)  a implies (force in + direction) - (force in - direction) = m2g - m1g = (m1 + m2) a. Solving for a gives the same result. Then, knowing a, T can be found by substitution. m1 m2 m1g T m2g

Atwood: Unit Check a = m2 – m1 m1 + m2 g kg - kg kg + kg m m s2 s2 units: = s2 s2 Whenever you derive a formula you should check to see if it gives the appropriate units. If not, you screwed up. If so, it doesn’t prove you’re right, but it’s a good way to check for errors. Remember, you can multiply or divide scalar quantities with different units, but you can only add or subtract quantities with the same units!

Atwood: Checking Extremes Besides units, you should also check a formula to see if what happens in extreme & special cases makes sense. m2 >> m1 : In this case, m1 is negligible compared to m2. If we let m1 = 0 in the formula, we get a = (m2 / m2 )g = g, which makes sense, since with only one mass, we have freefall. m2 << m1 : This time m2 is negligible compared to m1, and if we let m2 = 0 in the formula, we get a = (-m1 / m1 )g = -g, which is freefall in the negative (counterclockwise) direction. m2 = m1 : In this case we find a = 0 / (2m1)g = 0, which is what we would expect considering the device is balanced. Note: The masses in the last case can still move but only with constant velocity! m2 – m1 m1 + m2 a = g m1 m2