Physics 170 Week 5, Lecture 2 http://www.phas.ubc.ca/ gordonws/170 Physics 170 Week 5 Lecture 2 1
Textbook Chapter 5:Section 5.5-5.7 Physics 170 Week 5 Lecture 2 2
Learning Goals: Review the condition for equilibrium of a rigid body Consider the special case of two-force and three-force members Learn how to recognize and use what we learn about two-force and three-force members in the context of an example. Physics 170 Week 5 Lecture 2 3
Review: Equilibrium of a rigid body Consider a static rigid body with a number of forces F 1,..., F k acting on it at points r 1,..., r k and a number of couple moments M 1,... M q. The conditions for equilibrium are k F i = 0 (1) i=1 q M j + k M Oi = q M j + k ( r i r O ) F i = 0 (2) j=1 i=1 j=1 i=1 M Oi are moments due to forces, taken about a point O. The sum over moments are independent of the position of the point O. Furthermore, forces are sliding vectors they can be moved along their lines of action. Physics 170 Week 5 Lecture 2 4
Two-force members Theorem: Consider a subsystem of a larger system which is in equilibrium and which has only two forces and no couple moments acting on it. If that subsystem is to be in equilibrium, the forces must have equal magnitude, opposite directions and they must have the same line of action. Proof: Since we must have i F i = 0 and there are only two forces, F 1 + F 2 = 0 F 2 = F 1 Furthermore, the sum of moments must vanish. If F 1 acts at r 1 and F 2 acts at r 2, the sum of moments about O located at r O is ( r 1 r O ) F 1 + ( r 2 r O ) F 2 = ( r 1 r O ) F 1 ( r 2 r O ) F 1 = ( r 1 r O r 2 + r O ) F 1 = ( r 1 r 2 ) F 1 = 0 which can only be true if r 1 r 2 F 1. QED Physics 170 Week 5 Lecture 2 5
Example: The force on the pins at A and B must have equal magnitudes and opposite directions and they must have a common line of action. Physics 170 Week 5 Lecture 2 6
Three Force Members Theorem: If three non-parallel forces and no couple moments act on a body in equilibrium, the forces are concurrent, that is, their lines of action must have a common point of intersection. Physics 170 Week 5 Lecture 2 7
Proof: Consider the resultant force F 1+2 = F 1 + F 2. Together with F 3, they form a two-force system, i.e. F 1+2 = F 3 and (by the previous theorem) they have the same line of action, so can be taken as acting at the same point. Physics 170 Week 5 Lecture 2 8
F1 r1 ro F3 r2 F2 Moments computed about any point must vanish, so let us compute the moments about a point which is on the line of action of F 3 and therefore on the line of action of F 1 + F 1. ( r 1 r O ) F 1 +( r 2 r O ) F 2 = 0 ( r 1 r O ) F 1 = ( r 2 r O ) F 2 and, in particular, F 1, F 2,( r 1 r O ) and ( r 2 r O ) are all in the same plane. Thus, in particular, the lines of action of F 1 and F 2 must intersect (nonparallel lines in a plane must intersect). Physics 170 Week 5 Lecture 2 9
F1 r1 ro F3 r2 F2 It is a simple fact of geometry (parallelogram law of addition) that F 1, F 2 and F 1+2 lie in the same plane and that their lines of action intersect at the same point. Since F 3 = F 1+2 its line of action intersects the lines of action of F 1 and F 2 at that point. Physics 170 Week 5 Lecture 2 10
Example Consider the lever assembly. Find the direction of the reaction force at A. Physics 170 Week 5 Lecture 2 11
The short link is a two-force member The short link BD is a two-force member, so the resultant forces at pins D and B must be equal, opposite, and collinear. Although the magnitude of the force is unknown, the line of action must pass through B and D. Physics 170 Week 5 Lecture 2 12
The lever ABC is a three-force member The three nonparallel forces acting on it must be concurrent at O. The distance CO must be 0.5 m tan θ = 0.7m/0.4m θ = 60.3 Physics 170 Week 5 Lecture 2 13
Find the magnitude of the force at A: Physics 170 Week 5 Lecture 2 14
Describe the three forces: F C = 400Nî F = F ) (î + ĵ 2 ) F A = F A (cos θî + sin θĵ Physics 170 Week 5 Lecture 2 15
F C = 400Nî F = F ) (î + ĵ 2 ) F A = F A (cos θî + sin θĵ Decompose into components and apply equilibrium condition: i F xi = 0: F A cos θ F/ 2 + 400N = 0 i F yi = 0: F A sin θ F/ 2 = 0 The second equation implies F A = F 2 sin θ Plug this into the first equation: F cos θ 2 sin θ F 2 + 400N = 0 Physics 170 Week 5 Lecture 2 16
The equation which remains to be solved is which we can re-write as F cos θ 2 sin θ F 2 + 400N = 0 F = ( 2)(400) N 1 cot θ = ( 2)(400) N 1 (.4)/(.7) = (.7)( 2)(400) N (.3) where we recalled the result of the previous computation that tan θ =.7/.4. Finally, F = 1320 N Physics 170 Week 5 Lecture 2 17
Example: Find the angle of the ladder θ at equilibrium (so that the ladder will stay there without friction). Physics 170 Week 5 Lecture 2 18
Strategy for finding a solution: We recognize that the ladder is a three-force member. This means that the three forces acting on it must be concurrent. We can then solve the problem by finding the angle θ such that the three forces are indeed concurrent. We will assume that the reaction forces at the roof and wall are orthogonal to the roof and the wall, respectively. Physics 170 Week 5 Lecture 2 19
Free body diagram Physics 170 Week 5 Lecture 2 20
Geometry The vertical distance from the bottom to top of ladder is 18 sin θ. The horizontal distance from the bottom of the ladder to its center of gravity is 9 cos θ. The tangent of the angle between the normal to the roof and the horizontal, i.e. (90-40)=50 degrees, is equal to the ratio of these two distances, tan 50 = θ = 30.8 degrees 18 sin θ 9 cos θ = 2 tan θ θ = arctan 1 2 tan 50 Physics 170 Week 5 Lecture 2 21
Enrichment problem: Assume that the ladder is in equilibrium, that is θ = 30.8 degrees. Is this configuration stable? That is, if you change the angle a little bit, does the ladder tend to move back to its equilibrium position or does it tend to move away even farther from equilibrium? Physics 170 Week 5 Lecture 2 22
For the next lecture, please read Textbook Chapter 8:Section 8.1-8.2 Physics 170 Week 5 Lecture 2 23