a 10.0m long bar is attached by a frictionless hing to a wall and held horizontally by a rope that makes an angle of 53 deg with the bar the bar is uniform and weights 39.9 N. how far from the hinge should a 10.0kg mass be suspended for the tension T in the rope to be 125 N
Forfit to add the answer is 8.15m
a uniform beam LM has a weight of 60N and is 5 M long. it rest on a pivot at P. PM=1.6m. A 80N weight hangs from L.A vertical rope attached midway between P and M prevents the beam from rotating. calculate the tension T in the rope?
It's not 12.4, you have to find the perpendicular distance from T and the hinge. Also the bar is only 10 m so an answer longer than that makes no sense
To determine the distance from the hinge at which the 10.0kg mass should be suspended, we need to consider the torque balance in the system.
The torque equation is given by:
τ = r * F * sin(θ)
Where:
τ = Torque
r = Distance from the hinge to the point where the force is applied
F = Force
θ = Angle between the line of action of the force and the line connecting the hinge and the force
In this case, the torque due to the weight of the bar (39.9N) is counteracted by the torque due to the tension in the rope (125N) and the weight of the mass (10.0kg). Since the bar is held horizontally, the torque due to its weight can be calculated as follows:
τ_bar = r_bar * F_bar * sin(90°)
Since the bar is held horizontally, sin(90°) = 1. Therefore, the torque due to the bar's weight can be simplified to:
τ_bar = r_bar * F_bar
Similarly, the torque due to the tension in the rope and the weight of the mass can be calculated as follows:
τ_tension = r_tension * F_tension * sin(θ)
Substituting the given values into the torque equation, we have:
τ_tension = r_tension * 125N * sin(53°)
τ_mass = r_mass * 10.0kg * 9.8m/s² * sin(90°)
Since the system is in equilibrium, the sum of the torques is equal to zero:
τ_bar + τ_tension + τ_mass = 0
Substituting the calculated torques into the equation, we can solve for the unknown distance r_tension:
r_bar * F_bar + r_tension * 125N * sin(53°) + r_mass * 10.0kg * 9.8m/s² * sin(90°) = 0
Substituting the known values:
10.0m * 39.9N + r_tension * 125N * sin(53°) + r_mass * 10.0kg * 9.8m/s² * 1 = 0
Simplifying the equation:
399N + r_tension * 125N * sin(53°) + 98N * r_mass = 0
We can now solve for r_tension:
r_tension * 125N * sin(53°) = - (399N + 98N * r_mass)
r_tension = - (399N + 98N * r_mass) / (125N * sin(53°))
Now, substitute the mass of the weight (10.0kg) into the equation and calculate r_tension:
r_tension = - (399N + 98N * 10.0kg) / (125N * sin(53°))
This will give us the distance from the hinge at which the 10.0kg weight should be suspended for the tension in the rope to be 125N.