Question

5. A right triangular wedge of mass M and angle theta, supporting a cubical block of mass m on its side, rests on a horizontal table, as shown. a. What horizontal acceleration a must M have relative to the table to keep m stationary relative to the wedge, assuming frictionless contacts? b. What horizontal force F must be applied to the wedge to achieve this result, assuming a frictionless table top? c. Suppose no horizontal force is supplied to M and both surfaces are frictionless. Find the resulting acceleration of each mass relative to the table

Answer 1

PART A and PART B are simple

So, ma cos theta = mg sin theta to keep the block at rest or = a = g tan theta For a = g tan theta F = (g tan theta) (M + m) \r\n a_m Vector = a_m Vector, M + a_M a_m M Vector = (a cos theta i^hat - a sin theta j^hat) a_M Vector = -A i^hat a_m Vector = (a cos theta i^hat - a sin theta j^hat) - A i^hat For M N sin theta = MA = > N = MA/sin theta - ii For m mg - N cos theta = ma sin theta or mg - (MA/sin theta) cos theta = ma sin theta or mg sin theta - MA cos theta = ma sin^2 theta - iii N sin theta = m (a cos theta - A) N sin theta + ma = ma cos theta \r\n Now from ii (mA/sin theta) sin theta + mA = ma cos theta or MA + mA = ma cos theta
now PART SEE

**last line of 2nd page its Nsin@+mA=macos@

A is being written as a so correct that