A 4-m long, 150-kg steel beam is attached to a wall with one end connected to a hinge that allows the beam to rotate up and down. The other end of the beam is held in a horizontal position with a cable that makes a 27° angle with the beam and is attached to the wall. A mass of 75 kg is hung from the beam 3 meters away from the hinge (see (Figure 2)). What is the vertical component of the force that the hinge exerts on the beam?
(mass of steel beam) m = 150kg
(mass of extra thing) M = 75kg
(use the tension found in the previous part) F_t = 2833.2N
net force in y direction = ma
H_y + F_t * sin(27) - M * g - m * g = 0
H_y = 918.75N
A 4-m long, 150-kg steel beam is attached to a wall with one end connected to a hinge that allows the beam to rotate up and down. The other end of the beam is held in a horizontal position with a cable that makes a 27° angle with the beam and is attached
A 4-m long, 150-kg steel beam is attached to a wall with one end connected to a hinge that allows the beam to rotate up and down. The other end of the beam is held in a horizontal position with a cable that makes a 27° angle with the beam and is attached to the wall. A mass of 75 kg is hung from the beam 3 meters away from the hinge (see (Figure 2)). Now what is the tension force...
A 4-m long, 150-kg steel beam is attached to a wall with one end connected to a hinge that allows the beam to rotate up and down. The other end of the beam is held in a horizontal position with a cable that makes a 27° angle with the beam and is attached to the wall (see (Figure 1)). What is the tension force that keeps this beam in static equilibrium?
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