A cylinder and a wedge of same masses with a vertical face, touching each other, move along two
smooth inclined planes forming the some angle and ß respectively with the horizontal. Determine the
force of normal N (in newton) exerted by the wedge on the cylinder, neglecting the friction between
them. If m= kg, a = 60°, B = 30° and g = 10m/s2
Answers
Answer:
Explanation:
The cylinder is followed up on by the power of gravity m1g, the typical reaction N1 of the left inclined plane, and the ordinary reaction N3 of the wedge (force N3 has the horizontal direction). We will compose the equation of movement of the cylinder as far as projections on the x1-axis coordinated along the left inclined plane (Fig.):
m1a1 = m1g sin α - N3 cos α , (1)
where a1 is the projection of the speeding up of the cylinder on the x1 - axis.
The wedge is followed up on by the power of gravity m2g, the ordinary reaction N2 of the right inclined plane, and the typical reaction of the cylinder, which, as indicated by Newton's third law, is equivalent to - N3. We will compose the condition of movement of the wedge as far as projections on the x1-axis coordinated along the right inclined plane:
m2a2 = m2g sin α + N3 cos α. (2)
During its movement, the wedge is in contact with the cylinder. In this way, if the displacement of the wedge along the x1-axis is Δx, the focal point of the cylinder (together with the vertical essence of the wedge) will be uprooted along the horizontal by Δx cos a. The focal point of the cylinder will be in this manner displaced along the left inclined plane (x1-axis) by Δx. This implies during the procedure of movement of the wedge and the cylinder, the connection
a1 = a2 (3)
is fulfilled.
Comprehending Eqs. (1) - (3) at the same time, we conclude the power of typical weight N = N3 applied by the wedge on the cylinder:
N3 = 2m1m2/m1 + m2 tan alpha
Answer:
Explanation: