F = 0. x o F = -k x o v = 0 F = 0. F = k x o v = 0 F = 0. x = 0 F = 0. F = -k x 1. PHYSICS 151 Notes for Online Lecture 2.4.

Transcription

1 PHYSICS 151 Notes for Online Lecture.4 Springs, Strings, Pulleys, and Connected Objects Hook s Law F = 0 F = -k x 1 x = 0 x = x 1 Let s start with a horizontal spring, resting on a frictionless table. We pick a reference point on the ass for exaple, the center of the ass. he position of the center of the ass when the spring is unstretched is called the equilibriu point (x = 0). Now I pull the spring an arbitrary distance x to the right. he spring exerts a force in the direction opposite the displaceent (to the left in this case). he force is given by Hooke s Law: F = kx where k is the spring constant and has units of N/. If I pull the spring to the right, the spring exerts a force to the left. Alternately, I can push the spring in a distance x. Now the spring exerts a force toward the right. Reeber that Hooke s law only works when the displaceents are sall. If you ake a very large displaceent, Hooke s law doesn t apply anyore and none of what I about to tell you will apply either. F = 0 x o F = -k x o v = 0 F = 0 v =v o F = k x o v = 0 x o F = 0 v = v o Lecture 15 Page 1

2 Ex. 1 A spring with a force constant of 10 N/ is used to push a 0.7 kg block of wood against a wall, as shown. (a) Find the iniu copression of the spring needed to keep the block fro falling given that the coefficient of static friction between the block and the wall is (b) Does your answer to part (a) change if the ass of the block of wood is doubled? (a) F = 0 y fs g = 0 µ sfs g = 0 µ s ( kx) g = 0 g x = µ k s ( ) (0.7 kg) 9.81 s x = = 4.8 c (0.46) 10 N ( ) (b) Yes; the spring displaceent is proportional to the block s ass. Strings - in a taut string tension pulls equally to the right and left. If the string were cut, tension is the force that would be required to hold the string together. Lecture 15 Page

3 Pulleys - strings are often used in conjunction with pulleys. A string has the sae tension throughout its length and the pulley siply serves to redirect the tension. Ex. What is the stretching force applied to the broken lib? he tension in the rope is equal to the weight of the ass = (.50 kg) (9.8 /s) = 4.5 N. hus, suing forces about the pulley at the foot yields: Σ F = cos 30 + cos 30 F = 0 x 1 F = cos 30 = (.50 kg) 9.81 cos 30 = 4.5N s Statics is the analysis of situations in which there is no acceleration. here are any applications, including buildings, people's bones when they are in traction, and ore. Statics is a fancy way of saying: r F = 0 Reeber though, that this is actually two equations, because it is a vector equation: F F BOH of these conditions ust be true to have static equilibriu. Note also that we are talking only about translational equilibriu we haven t discussed rotation yet x y = 0 = 0 Ex. 13-3: A 0.0 kg sign hangs as shown. What are the tensions in the two wires? he first thing to do is to identify a point that is in static equilibriu. his point ust have the forces that are unknown acting on it. We will choose the point at which the strings eet the sign. he free-body diagra will look like: Lecture 15 Page 3

4 g We know we will have to work with both coponents (x and y), so the first thing to do is to break each of the tensions into coponents. Reeber that for strings, beas, etc., the tension will always be along the string. 1x = - 1 sin(50) 1y = - 1 cos(50) x = sin(50) y = - cos(50) Now apply Newton's second law in both the x and the y directions. Reeber that the acceleration in all directions ust be zero. F x = 0 F y = 0 1sin( 50) + sin( 50) = 0 1cos( 50) + cos( 50) g = 0 1sin( 50) = sin( 50) he equation in the x direction tells us that 1 = We ight have already guessed this, because the two angles are equal. Use this in the y- equation. Call 1 = = cos( 50) = g g = cos( 50) d ( 0. 0kg) s = = 15 N cos( 50) i Ex. 4: A fish of ass is suspended as shown. he string AB will break when the tension exceeds 10.0 N. What is the axiu ass fish that can be supported? Start by drawing a free body diagra. Since we want to learn soething about the tension in AB, we need to draw the FBD for a point on which that force acts. In this case, point B is the best 35 point. First, find all of the B coponents 1 of the forces. A B 35 C g Lecture 15 Page 4

5 1x = - 1 1y = 0 x = cos (35) y = sin (35) hen apply Newton's second law for the case of a = 0. F x = 0 F y = 0 1+ cos( 35) = 0 sin( 35) g = 0 1 = cos( 35) 1 = cos( 35) N = cos( 35) 1. N = he axiu that 1 can be is 10.0 N, so Now use this in the y-equations sin( 35) = g sin( 35) = g b1. Ng sin( 35) = 98. s kg = You ry It! he syste shown is in equilibriu. (a) Find the frictional force exerted on block A given that the ass of block A is 8.50 kg, the ass of block B is.5 kg, and the coefficient of static friction between block A and the surface on which it rests if (b) If the ass of block A is doubled, does the frictional force exerted on it increase, decrease, or stat the sae? (a) Let be the tension in the slanting stretch of rope. hen sin 45 is the tension in the rope supporting ass B, and cos 45 is the tension in the rope pulling on ass A. But sin 45 = cos 45, and so fs on A = cos 45 = sin 45 = Bg = (.5kg) 9.81 =.1 N s Lecture 15 Page 5

6 which is below fs,ax = µ sa g = 0.30(8.50 kg) 9.81 = 6.7 N. s (b) So long as ass A is heavy enough for fs, ax.1 N, fs is not affected by changes in ass A; f s stays the sae. Connected Objects hese are probles with two asses that are connected by a rope over a pulley. Both asses have the sae acceleration. It is necessary to draw a FBD for both asses. he coordinate systes ust be related so that a positive acceleration of one ass is a positive acceleration for the other. Atwood s Machine We will assue here that M > and that pulley is frictionless. Our goal is ost often to deterine the acceleration and only occasionally to deterine the tension. Note that our first goal is to draw FBDs for both asses and that they ust have different coordinate systes. he coordinate systes ust be set up so that a positive acceleration for is oving upward while a positive acceleration for M is oving downward. Σ Fx = g = a Σ Fx M= Mg = Ma Σ Fx = g = a Σ F = Mg = Ma x M Mg g = a + Ma gm ( ) = a ( + M) ( M ) a = g ( + M) We can then substitute our expression for a back into either equation to solve for. You should do this and verify the result given below. M ( + M) Lecture 15 Page 6 = g

7 Horizontal Atwood s Machine N 1 g g +x Σ F = = a x 1 Σ F = g = a x 1 ( + ) ( ) g= a+ a= + a a = 1 1 g 1 Horizontal Atwood s Machine (with Friction) Σ F = F = a x 1 Σ F = g = a x f 1 ( ) g F = a+ a= + a f 1 1 g F g µ g µ a = = = f k 1 k 1 ( + ) ( + ) ( + ) g g +x F f N 1 g he asses ay or ay not accelerate depending upon the values of the asses and coefficient of friction. If > µ s 1 then the asses will accelerate with the expression given above. If not, then acceleration is equal to zero. Lecture 15 Page 7

8 Lecture 15 Page 8