2. 1. Equilibrium of rigid body In that figure, the stem doesn’t move/rotate. The stem is in equilibrium. The terms to get this equilibrium are: 1. Wa=weight a, Wb=weight b, N=normal force. 2. t=torque N a b
3. Don’t forget that torque is the multiplication between force and radius (t=FxR) In this case, F=W, so we can change the torque formula to be:
4. Example: In that figure,the stem is in equilibrium. the fulcrum is in the centre of the stem. Known that Rb(the radius of ‘b’ from the fulcrum)=2cm and the length of stem is 4cm. The weight of a is 60 kg and b is 30 kg. determine the radius of ‘a’ from the fulcrum! a b
5. Answer: You can answer it??? From our terms, so let we use the second term. t=0! Note: g=gravitational acceleration
6. Do it !!! A cube is hung as in figure and the stem is tied by rope as in figure.the angle formed is 53.Determine the string tension of T to make the system equilibrium!(the mass of stem is ignored) T? 53 L1=1m L2=2m m=5kg
7. 2.A homogenous stem as in figure, the mass stem is 8 kg, and known that tg A= ¾. Try to determine the string tension T and coefficient of friction(f)!(g=10 m/s^2) T f A 1m 4m m=4kg
8. 3.A cube in the bench has 40kg and is put as in figure. Determine the normal force of A and B! (the mass of bench is ignored and g=10) 6m 2m A B
9. conclusion: from our questions above, we can get the conclusion that in the equilibrium system has two terms(F=0 and t=0) And don’t forget that force is always perpendicular to the radius(the length of a bar from fulcrum to the force)
11. EMPHASIS Emphasis is located in symmetrical line: exp: square,rectangle,cylinder,etc. emphasis in sharp object: In triangle: has a formula to determine the emphasis h=high of triangle. In cone: also has a formula
12. 3.Emphasis in semicircle and half solid ball: semicircle: half solid ball: 4.Emphasis in combination object: For number four, we will learn it more to the next meeting(new slide)