A 4.5 m long, non-uniform beam weighing 750 N, supported by two weightless cable

Question

A 4.5 m long, non-uniform beam weighing 750 N, supported by two weightless cables, is at rest in horizontal position, as shown. The cable on the right has a tension of T_1 = 375 3 N. (a) The location of the center of gravity of the beam measured from its left end is _____ (b) The tension T_2 in the other cable is _____ A non-uniform, 2.0 m long ladder weighing 400 N leans against a smooth vertical wall (no friction). The ladder makes an angle of 30 degree with the vertical. The coefficient of static friction at the ground is 0.3. A child with weight of 300 N can climb the ladder of 1.2 m before it starts to slide down. (a) The distance of the center of gravity of the ladder from the contact point in the ground is _____ (b) The normal force coming from the smooth wall is _____ Starting from rest, a 1.0 m diameter, uniform cylindrical wheel (I=M R^2) is rotating under a constant angular acceleration of 3.0 rad/s^2. (a) The magnitude of the centripetal acceleration of a point on the rim of the wheel at t = 4.0 s is _____ (b) If the mass of the wheel is 40.0 kg, its angular momentum at t = 4.0 s is _____ A 4.0 kg block, lying on a smooth table, is connected to a 5.0 kg block by means of a weightless cord passing over, without slipping, a cylindrical pulley of 2.0 kg mass and 10.0 cm radius placed at the edge of the table. The system is released from rest at t = 0. (a) In 0.5s the 5.0 kg block will drop by _____ (b) The tension in the part of the cord connected horizontally to the 4.0 kg block is _____ A uniform circular wheel (I = M R^2) with radius, R = 1.0 m and mass, M = 500 kg is spinning at 120 rev/min. At a particular moment of time (that we denote t=0) the wheel is subjected to a frictional force F = 20.0 N applied tangentially at its rim. (a) The magnitude of the angular velocity of the wheel (in rad/s) after t = 25 s is _____ (b) The number of revolutions done by the wheel before coming to rest is _____

Explanation / Answer

If the system is in static equilibrium, the net torque must be zero.

WX=T1L

X= T1L/W

= (375)*sqrt(3)*4.5 / 750

= 3.89 m

b.

If the system is in static equilibrium, then the net force must be zero.

W+T1+T2=0

Then the tension is

T2=-W-T1

= -1399.52 N

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