A cylinder of weight w is resting on a fixed V-groove as shown in figure.Calculate normal reactions between the cylinder and two inclined walls.
(A) w,w
(B) w/2,w/2
(C) w,w/2
(D) w/2,w
Answers
cylinder in V groove.
Let us assume that the angles A made by inclined surface in groove with the vertical are both same. Let the cylinder rest symmetrically in the groove. Surface ST is horizontal. Normal reactions are N1 and N2. Also, frictions f1 and f2 are equal if the groove surface is of same kind on both sides.
The cylinder is in a static equilibrium. Balance forces in vertical direction and horizontal direction.
N1 cos A - f 1 Sin A - N2 Cos A + f2 sin A = 0
=> (N1- N2) Cos A = (f1-f2) Sin A --- (1) f1 <= mu N1 and f2 <= mu N2. Substituting that in (1), we get Tan A <= 1/mu N1 Sin A + N2 Sin A + f1 Cos A + f2 Cos A = W (sin A + mu Cos A) (N1 + N2) = W --- (2)
moments of the forces wrt Q: SQ * N1 = QT * N2 => N1/ N2 = QT / SQ --- (3)
Substitute in (2) to get: N2 (1+ QT/SQ) = W / (sin A + mu Cos A) N2 = W *QS /[( sinA + mu Cos A) (Qs+QT) ] N1 = W * QT / [( sinA + mu Cos A) (Qs+QT) ]
If QS = QT then, N = W / [ 2 (sin A + mu Cos A) ]
hope it is helpful to you
