- #1

John O' Meara

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A 45 deg wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge find its acceleration. Clue if A=3g then [tex] \ddot{y} = g [/tex]

I'm not sure of the forces on the block, if A is parallel to the base of the wedge then what component of A acts on the block and what is its magnitude? I don't know. The force on the wedge is (M+m)*A, where M is the mass of the wedge. Noting that if A is large enough the block will stay put on the wedge as the wedge is accelerated away. The table is taken as the x-axis. [tex] \sum{F_x} = m*a_x [/tex]. Please see attached diagram in pdf format. Thanks in advance for the help.

I'm not sure of the forces on the block, if A is parallel to the base of the wedge then what component of A acts on the block and what is its magnitude? I don't know. The force on the wedge is (M+m)*A, where M is the mass of the wedge. Noting that if A is large enough the block will stay put on the wedge as the wedge is accelerated away. The table is taken as the x-axis. [tex] \sum{F_x} = m*a_x [/tex]. Please see attached diagram in pdf format. Thanks in advance for the help.