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4-4 Forces In Nature There are many types of forces: The Fundamental Forces / Field Forces The fundamental forces are all types of field forces which means “action at a distance” forces, or no direct contact forces. What that means is that as objects enter a field, they are subject to the effects of the field. In order of decreasing strength, they are: 1. Strong Nuclear Force = the attractive force that holds the subatomic particles of a nucleus together. You probably remember this one from AP Chemistry! 2. Electromagnetic Force = the force between electrical charges. May be a force of attraction or repulsion. It is assumed that even though the masses of an electron and proton are different (mass of an e- = 1/1837 of mass of a proton), their charges are equal in magnitude, but opposite in direction. 3. Weak Nuclear Force = the force between subatomic particles during certain types of radioactive decay. 4. Gravitational Force = the force of mutual attraction between two objects. Whether viewed from a Newtonian or an Einsteinian perspective, there is a force of gravity between any two objects. Both the strong and weak nuclear forces operate only on a nuclear scale, and are thus only viable within a range of 10-15 m. Beyond this range, they have no effect. Classical physics deals only with gravitational and electromagnetic forces which have, for intensive purposes, infinite range. Contact Forces Any force that involves direct contact between two or morr objects is a contact force. Some examples of contact forces are: A.) Solids 1. Push / pull – may be parallel, perpendicular, or anti-parallel (at an angle) to the contact surface. 2. Normal Force = any force perpendicular to the contact surface. These are particularly relevant in drawing free-body diagrams (FBD’s). 3. Frictional Force = the force (or component of the force) that is parallel to the contact surface and oppositely directed to the sliding motion of the object. Frictional forces are response forces. Their magnitudes increase to a certain point in direct response to the push / pull exerted on the object. We will cover this in greater detail next chapter. B) Springs When a spring is either compressed or stretched by a certain amount, x, the force it exerts back is given by: where k is the force constant measuring the stiffness of the spring. Q: Why is there a negative sign in the Hooke’s Law equation? A: The force the spring exerts back is in the opposite direction of the original force. Because it acts in the opposite direction, the effect is to return, or restore, the object to its “equilibrium” position, and is thus called a restoring force. Q: What does equilibrium mean? Describe the types of equilibrium. A: Equilibrium is when the sum of the forces, or the net force, = zero. The two types of equilibrium are: Static equilibrium – the object is at rest and any forces that exist are not enough to accelerate the object. Dynamic equilibrium – the sum of the net forces = 0; the object is not at rest, but it is not accelerating. In other words, it is moving at constant velocity. The natural resting place of a spring is called its stable equilibrium. In a molecule or solid, the forces between atoms varies approximately linearly with the change in separation (for small changes), like the forces in a spring. If you push or pull the object greatly, you can compress, warp, stretch, or even break it, depending on its internal structure. Clamping down on a plastic block produces stress patterns that are visible when viewed with polarized light. C) Strings Strings are used to pull things. Think of a string as being like a spring with such a large force constant that the extension of the string is negligible. Though strings can flex (out), or bend, they can’t be used to push things. The amount of force that, technically, one segment of string exerts on the next is called tension. Really, tension is the amount of force in a spring, string, wire, or rope, et cetera. My daddy taught me that! D) Constraints A condition that restricts or controls the motion of another object in any way is called a constraint. For example, a railroad car moving along a track; a wooden pony on a carousel; a bobsled on a luge track. Before delving fully into the next example and the next section, let’s review some notes from 10th grade on free-body diagrams (FBD’s). If ∑ F = 0, the object is not accelerating. Which is / are accelerating below? Applied Forces Net Force DEF: Free-body Diagram (FBD) = a basic diagram that: Considers only one object at a time Represents the object with a simple sketch or dot uses vectors to represent all relevant forces acting on that object Superimposes a coordinate system to give direction and magnitude to the vectors NOTE: All forces must be “like” in that they contribute to the net motion of the object NOTE: As a matter of convention, start all vectors from the center of the object Draw Free-Body Diagrams (FBD’s) for all of the following: Ex 1) A block sitting on a table Ex 2) A piece of rope being pulled on by two equally strong children Ex 3) A child on a stationary sled pointed down a hill (300 incline) Ex 4) A child sledding at constant velocity down a hill (300 incline) Ex 5) This guy hanging from the ceiling wires. (C’mon, it could happen!) You try it. 1. A person sliding down a slide (450 incline) 2. A crate being pulled across the floor 3. A hockey puck sliding across the ice 4. Two people successfully pushing a stalled car 5. A ball rolling down a frictionless ramp See Ex 4-5, p 94. A 110 kg basketball player hangs on the rim following a slam dunk. Prior to dropping to the floor, he hangs motionless with the front of the rim deflected down a distance of 15 cm. Assume the rim can be approximated by a spring and calculate the force constant, k. This diagram is appropriate because it says he hangs motionle means he is not accelerating in any direction. Because of that, t are all balanced. To say he is hanging there implies the forces the y-direction. Also, they told us to assume the rim acts like which means that force will be calculated using Hooke’s Law. The sum of the forces will be the weight, in Newtons, of the basketball player + the spring force of the rim. Q: What units will you end up with if you divide N by kg? Don’t forget to convert cm to m. ∑Fy = wy + Fy-rim = ma mg + -kΔy = m 0 mg = kΔy mg = k Δy k = (110 kg)(9.81 N/kg) = 7.19 x 103 N/m 0.15 m 4-5 Problem Solving: FBD’s From your text: See Ex 4-6, p 96 During your winter break, you enter a dogsled race in which students replace the dogs. Wearing cleats for traction, you begin the race by pulling on a rope attached to the sled with a force of 150N at 250 to the horizontal. The mass of the sled-passenger-rope object is 80 kg and there is negligible friction between the sled runners and the ice. Find: a) the acceleration of the sled b) the normal force, Fn, exerted by the surface on the sled. What did they do to the vectors? What does the black vector, m a , signify? There are three vectors to consider here. Their sum will be equal to ma. Conceptually: The normal force has to cancel out the difference between the y component of the F and the weight, because Fnet is in the x-direction only! a) Because ma is in the x-direction only, and weight and normal are in the ydirection only, we have: Fn + w + F = ma This gives us two sets of equations to deal with: x and y directions Fn,x + wx + Fx = max AND 0 + 0 + (150 N)(cos 250) = 80 kg ax m0 ax = (150 N)(cos 250) 80 kg 0 – (150 N)(sin 25 ) ax = 1.70 m/s2 b) Fn,y + wy + Fy = may Fn,y + -mg + Fsin = Fn = mg - Fsin = (80 kg)(9.81 N/kg) = 721 N See Ex 4-7, p 97 You are working for a big delivery company, and must unload a large, fragile package from the truck using a delivery ramp. If the downward component of the velocity of the package when it reaches the bottom of the ramp is greater than 2.5 m/s (the speed it would have if you dropped it from a height of 1 ft), the package would break. What is the largest angle at which you can safely unload? The ramp is 1 m high, has rollers (i.e., is frictionless), and is inclined at an angle θ. This FBD shows the x-axis along th direction of natural motion, o acceleration, as recommended. Note tha the weight vector always points straigh down, and that the Fn is always th contact surface, or x-axis, in this case. Some very important “tricks” for solving these types of problems: 1. The weight vector is straight down and = - mg. Does that make sense in terms of N2? (F = ma → F = m(-g)). 2. All of the θ’s are equal. (Do you know why? θ1 = θ2 Alt Int ’s are equal; θ2 = θ3 complements of the same are equal ) Because that’s true, θ will actually be resolved along the y-axis. Look at it! For that reason, the weight vector, which needs to be resolved in both x and y will switch wrt x and y. So, the weight vector resolved in x will = mgsin θ, and resolved in y will be mgcos θ. 3. Drawn this way, the normal force will always be only in the + y-direction. Now for the problem-solving process: 1st: Find the sum of the forces in x, and if the object will accelerate in x, set them equal to ma. Fn,x = 0. Fx : 0 + mgsin θ = max ax = gsin θ The m’s cancel We don’t know ax, but we do know vf and vi, so we can use those to set up and solve for vf in terms of ax vf2 = vi2 + 2axx vf2 = 2axx Since vi = 0 vf2 = 2 gsin θ x Substituting ax from above For this next part draw a couple of right triangles, and use right-triangle geometry. x h Ө When x = the length of the ramp (as it’s moving along the x-axis), h = x sin Ө. Thus, vf2 = 2gsin becomes vf2 = 2gh vf vd Ө Looking at the velocity in the downward direction, vd, will give: vd = vf sin Ө Solve the equation in the box above for vf, and you get vf = 2 gh . Substitute, and you get = 2 gh sin Ө equation for vd. Note: There is an additional sin Ө from the 2.5 m/s = 2(9.81m / s 2 )(1.0m) sin Ө Ө = 34.40 See Ex 4-8, p 98 A picture weighing 8N is suspended by two wires with tensions T1 and T2 as shown. Find each tension. Remember that tension just means the force in a string, spring, rope or wire. The 600 angle on the top = the 600 angle on the bottom because alt int ’s are equal. Same is true for the 300 ’s. The picture is stationary; that means there is no acceleration in any direction. ∑Fx: - T2cos 600 + T1 cos 300 = 0 T2 = T1 cos 300 = 1.73 T1 cos 600 ∑Fy: T2sin 600 + T1 sin 300 – w = 0 1.73 T1 sin 600 + T1 sin 300 – w = 0 evaluate substitute for T2 add w to both sides, the products, sin’s, and cos’ 1.5 T1 + 0.5 T1 = 8N 2 T1 = 8N T1 = 4N T2 = 6.93N (7N) See Ex 4-9, p 99 As your jet plane speeds down the runway on takeoff, you decide to determine its acceleration, so you take out your yo-yo and note that when you suspend it, the string makes an angle of 220 with the vertical. a) What is the acceleration of the plane? b) If the mass of the yo-yo is 40 g, what is the tension in the string? Ө Ө Draw the y-axis, and then another line parallel to the yaxis. Then, these two Ө’s are equal. (Alt int ’s are ) So, they used the fact that alt int ’s are , and then moved the tension vector up to the origin. The way this is drawn, T will be resolved along the y-axis (not the x-axis). This means x will be in terms of sin Ө, and y. will be in terms of cos Ө. Because of how the acceleration vector is drawn, we know there is only acceleration in the x-direction. ∑Fx: T sin Ө = max There is an acceleration in x ∑Fy: 2nd eq T cos Ө - mg = 0 There is not an acceleration in y. Use this to solve for T. T cos Ө = mg T = mg = (0.040 g)(9.81 m/s2) = 0.423 N Now, subst. into the st 1 eq cos Ө cos 220 to solve for ax ax = (0.423 N)(sin 220) = 3.96 m/s2 0.040 kg See Ex 4-10, p 100 Suppose that your mass is 80 kg, and you are standing on a scale fastened to the floor of an elevator. The scale measures force and is calibrated in N’s. What does the scale read when (a) The elevator is rising with upward acceleration of magnitude a; (b) the elevator is descending with a downward acceleration of magnitude a’; (c) the elevator is rising at 20 m/s and its speed is decreasing at a rate of 8 m/s2? Remember this from 11th grade? Draw FBD’s for each scenario. All the motion is in the y-direction only, and is some derivation of the FBD below. (a) ∑Fy: Fn,y + wy = may Fn,y – mg = may Fn,y = mg + may Fn,y = m(g + ay) This will make you weigh morr because you are adding something to “g” (b) ∑Fy: Fn,y + wy = ma’ Fn,y – mg = m(- a’) Fn,y = mg - ma’ Fn,y = m(g – a’) This will make you weigh less because you are subtracting something from “g” (c) Apply ∑Fy: Fn,y + wy = may Fn,y – mg = may Fn,y = mg + may Fn,y = m(g + ay) = = (80 kg)(9.81 m/s2 + - 8.00 m/s2) = 145 N the So, you weigh much less, as you should; acceleration is almost g. 80 kg = 784.8 N. 145 / 784.8 18.5 % 9.81 – 8.00 = 1.81 1.81 / 9.81 18.5 %. The percentages are the same, as they should be!