Now the table is tilted at an angle of theta = 80.0 degree with respect to the v

Question

Now the table is tilted at an angle of theta = 80.0 degree with respect to the vertical. Find the magnitude of the new acceleration of block 1. m/s2 At what "critical" angle will the blocks NOT accelerate at all? Now the angle is decreased past the "critical" angle so the system accelerates in the opposite direction. If theta = 35.0 degree find the magnitude of the acceleration. m/s2 Compare the tension in the string in each of the above cases on the incline: T theta at 80.0 degree = T theta critical = T theta at 35.0 degree T theta at 80.0 degree > T theta critical > T theta at 35.0 degree T theta at 80.0 degree

Explanation / Answer

3) forces acting on m1 = m1gcos(80) at an angle 80 degrees to vertical and tension T = m2g acting parallel to first force but in opposite direction. so, m1a = m2g - m1gcos(80) so, a = g(m2 - m1cos(80))/m1 m/s2 Ans. 4) the blocks will not move if a=0 so, m2g = m1gcos(q) q = cos inverse m2/m1 Ans. 5)

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