Equilibrium of Rigid Bodies: Torque and Newton's Laws - Prof. Braden Abbott, Study notes of Physics

Download Equilibrium of Rigid Bodies: Torque and Newton's Laws - Prof. Braden Abbott and more Study notes Physics in PDF only on Docsity! Physics 2414 Group Exercise 12 Name 1: OUID 1: Name 2: OUID 2: Name 3: OUID 3: Name 4: OUID 4: So lu tio ns So lu tio ns Section Number: Equilibrium of rigid bodies Torque: Torque about an axis due to a force ~F is determined by the expression τ = rF⊥ = r|~F|| sin θ| (1) where r is the distance between the axis and the point of action of the force, and F⊥ is the component of the force that contributes to the rotation. In the diagram shown F⊥ = |~F|| sin θ|. (2) θ ~F F⊥ F|| r Newton’s second law for translational motion of a mass m is ~Fnet = m~a (3) where ~Fnet is the sum of all the forces acting on the mass, and ~a is the translational acceleration of the mass. Newton’s second law for rotational motion of a rigid body with rotational inertia I about an axis is τnet = Iα (4) where τnet is the sum of all the torques acting on the rigid body, and α is the angular acceleration of the rigid body about the axis. 1 Problems A ladder of length l = 20 meters weighing mLg = 500 Newtons rests against a wall at a point h = 12 meters above the ground. The center of mass of the ladder is at the center of the ladder. A man weighing mpg = 800 Newtons climbs a distance x = 15 meters up the ladder. The friction on the floor keeps the ladder from slipping. Assume that the wall is frictionless. 15 m 12 m mp~g mL~g ~Ff ~Nv ~Nh θ A B 1. Newton’s law for translational equilibrium: The Newton’s law for the translational equilibrium in our problem is described by the vector equation ~Fnet = m~a (equilibrium ⇒ ~a = 0) (5) ~Nh + mp~g + mL~g + ~Ff + ~Nv = 0 (6) (a) Write down the y-component of eqn. (6) and thus determine the magnitude of the vertical normal force ~Nv. 0 − mpg − mLg + 0 + Nv = 0 Nv = mpg + mLg = 800 + 500 = 1300 Newtons (7) 2