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 that keeps this beam in static equilibrium?
(mass of steel beam) m = 150kg
(mass of extra thing) M = 75kg
(where the extra mass thing hangs) x = 3m
net τ = τ_cable - τ_gravity - τ_mass = 0
τ_gravity = m * g * (L / 2) note: L/2 b/c that's where center of gravity is for beam
τ_cable = F_t * sin(27 degrees) * L
τ_mass = M * g * x
F_t * sin(27) * 4 - 150 * 9.8 * (4/2) - 75 * 9.8 * 3= 0
F_t = 2833.2N
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)). What is the vertical component of...
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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