Question

CLASS ID#: (2) GIVEN (35 Prs): A uniform block cabinet of mass m and effective dimensions shown is mounted on small casters of negligible mass and diameter which allow it to move freely (u=0) on the floor. A horizontal force of F is applied at a distances above the base of the cabinet FIND: (A) (-10 Pts) The ACCELERATION of the cabinet: (B) (-15 Pts) The RANGE OF S VALUES for which the cabinet will not tip forward or backwards. (©) (-10 Pts) The ROLLER REACTION(s) as the cabinet reaches its limit on tipping forward. ME3133: Dynamics Spring 2018 - Final Exam

Answer 1

FBD: From Newton's 2nd law of motion, < F = ma I F * > la F т mg TEL No, N2 - Normal reactions at Casters. 1 N N2 (6 ) S= mgh 2 F Consider tipping forward ie. Ng = 0 (limiting Case) Moment about point B, Mg = (limiting case) F.5 - mg. Ķ=0 back Consider tipping forward N; -o limiting Case) Moment about point A, MA = 0 (limiting Case) F. 9 + mg. Ź - mgt - mgL mgh · Range of values of s i.e. -0. S = 2F are 2F 2 F But of s are negative values physically not clear from o to mal for not to tip. So S Varies 2F

Vertical is no acce keration in (c) As there direction, we have Nit N₂ = mg Ny = 0 For Case of forward tipping, Ny = ng
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