PHYSICS 111 Rotational Dynamics Torque EQUATIONS Torque t

PHYSICS 111 Rotational Dynamics Torque

EQUATIONS: Torque (t) = F * L �T=F * d sin ɵ If not rotation. Delta Fy: 0 Delta Fx: 0 Torque : 0 For uniform distribution of mass Weight is @. 5 L Summation of t = I ᾰ I = Summation of mr^2

PROBLEM A meter stick pivoted at the 50 cm mark is subject to two forces, as shown below. The 10 N force is applied to the 0 cm mark and the 15 N force is applied at the 75 cm mark. Which way will the meter stick rotate.

SOLUTION: SO the summation of forces is the F y direction have to equal zero And we want to know the summation of t Fy = F fulcrum + -15 + - 10 = �F fulcrum = 25 N T = F* L T = (0 * -10 ) + (25 * 25) + (75 * -15) t= 1250 + - 1125 = 125 N * m Positive means it is turning ccw

PRACTICE PROBLEM 1: A string is tied to a doorknob 0. 23 m from the hinge. The force applied to the string is 25 N. What is the torque about the hinge?

SOLUTION: T =F*L �F= 25 N �L =. 23 T=F * L T = 25 * (. 23 sin 57) t = 25 *. 1003 = 2. 507 N* m

PRACTICE PROBLEM 2 A man whose weight is 345 N is poised at the right end of a diving board with length 7. 21 m. The board has negligible weight and is supported by a fulcrum 2. 40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board.

SOLUTION: Delta Fy: 0 = -345 + F fulcrum + -F bolt Delta Fx: 0 = No x forces Torque : 0 = Summation of F * L Considered Bolt axis point T=0= 0*f bolt + F ful (2. 40) + -345 (7. 21) F fulcrum = 1036 N F bolt = -691. 44 N

PRACTICE 3: Consider the system below for question #1: � Determine the location of the 30 N

SOLUTION: Summation of t =0 = Summation of L *F �(35 cm)(35 N) = (x cm) (30 N) �X = 85. 83

DETERMINE THE FORCE AT 80 CM

SOLUTION: X= T 3. 5 = 0 = F* L � 5 (35) + 4 (10) -2 (45) = (. 50)(5) + x (35)