Chapter 8 Rotational Equilibrium and Rotational dynamics Torque and Equilibrium First Condition of Equilibrium The net external force must be zero r Σ F = 0 or r r Σ Fx = 0 and Σ Fy = 0 This is a statement of translational equilibrium •The Second Condition of Equilibrium states r Στ = 0 • The net external torque must be zero A hobbit house Three Factors affect torque •The magnitude of the force •The position of the application of the force •The angle at which the force is applied τ = r F sin θ F sinθ τ = r F sin θ F cosθ Torque and Equilibrium First Condition of Equilibrium The net external force must be zero r Σ F = 0 or r r Σ Fx = 0 and Σ Fy = 0 This is a statement of translational equilibrium •The Second Condition of Equilibrium states r Στ = 0 • The net external torque must be zero Torque direction: Right hand rule again Force turns it in the counterclockwise direction Force turns it in the clockwise direction Center of Gravity In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at a single point Center of gravity xcg Σmi xi Σmi y i = and ycg = Σmi Σmi A few pointers. If a body has a symmetry and it has a uniform density then the cg is on the line of symmetry. The center of symmetry coincides with the cg. The cg might be outside the object Example Find the cg of a 4x8 uniform sheet of plywood with the upper right quadrant removed. m1 = 2 M ; ( x1 , y2 ) = ( 2,2) y 4 x a 4 m2 = M ; ( x2 , y2 ) = (6,1) xcg = m1 x1 + m2 x2 2M ⋅ 2 + M ⋅ 6 10 ⋅ M 10 = = = ft m1 + m2 2M + M 6⋅ M 3 ycg = m1 y1 + m2 y2 2M ⋅ 2 + M ⋅1 5 ⋅ M 5 = = = ft 2M + M 3⋅ M 3 m1 + m2 8 Torque and Equilibrium First Condition of Equilibrium The net external force must be zero r Σ F = 0 or r r Σ Fx = 0 and Σ Fy = 0 This is a statement of translational equilibrium •The Second Condition of Equilibrium states r Στ = 0 • The net external torque must be zero Example of a Free Body Diagram-Ladder 75N 75N free body diagram shows normal force and force of static friction acting on the ladder at the ground In-class quiz 18-1 Find the force P of the wall on the top of the 10 meter ladder that weights 75 N 40° A. B. C. D. E. 50 N 25N 30 N 21 N 45 N 75N 40° 75N In-class quiz 18-1 Find the force P of the wall on the top of the 10 meter ladder that weights 50 N 40° A. B. C. D. E. 50 N 25N 30 N 21 N 45 N 40° A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips? A. 42o B. 22o C. 18o D. 51o E. 39o A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.62. What minimum angle can the ladder make with the floor before it slips? A. 42o B. 22o C. 18o D. 51o E. 39o Example: a ladder against a wall What minimum angle can the ladder make with the floor before it slips? P ∑F = 0 ∑F = 0 ∑τ = 0 x y fs − P = 0 y x n − mg = 0 L PL sin θ − mg cos θ = 0 2 f s = µ s n = µ s mg L µ s mgL sin θ − mg cos θ = 0 2 N mg θ fs 1 1 µs sinθ − cosθ = 0 tan θ = 2µ s 2 Torque and Angular Acceleration Newton’s Second Law for a Rotating Object Στ = Iα analogous to ∑F = ma I = moment of inertia 2 2 I = Σ m r = MR For Uniform Ring i i moment of inertia depends on quantity of matter and its distribution and location of axis of rotation Other Moments of Inertia

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