A uniform steel beam of mass m1= 300.0 kg is held up by a steel cable that is connected to the beam a distance L= 5.0 m from the wall, at an angle θ= 40.0∘ as shown in the sketch. The beam is bolted to the wall with an unknown force F⃗ exerted by the wall on the beam. An object of mass m2= 50.0 kg resting on top of the beam, is placed a distance d= 4.0 m from the wall. For simplicity, assume the steel cable to be massless. Use g=9.8 m⋅s−2 for the gravitational acceleration.
(a) Find the tension (in Newton) in the cable. Start by drawing a free-body diagram for the beam, then find equations for static equilibrium for the beam (this will involve force equations and torque relations).
T=
(b) Find the horizontal and vertical components of the force (in Newton) that the wall exerts on the beam.
Fx=
Fy=
T=2903.27
Thanx.. Can you also give out the rest?
To find the tension in the cable, we can start by drawing a free-body diagram for the beam and analyzing the forces acting on it.
Let's consider the forces individually:
1. The weight of the beam (m1 * g) acts vertically downward.
2. The weight of the object on top of the beam (m2 * g) also acts vertically downward.
3. The tension in the cable acts at an angle θ with respect to the horizontal axis.
4. The force exerted by the wall (F) acts horizontally.
Now, let's analyze the forces in equilibrium along the horizontal axis (x-axis):
1. The horizontal component of the tension in the cable (T * cos(θ)) acts to the right.
2. The force exerted by the wall (F) acts to the left.
Since the beam is in static equilibrium, the sum of these forces along the x-axis must be zero:
T * cos(θ) - F = 0 -----(1)
Next, let's analyze the forces in equilibrium along the vertical axis (y-axis):
1. The vertical component of the tension in the cable (T * sin(θ)) acts upward.
2. The weight of the beam (m1 * g) acts downward.
3. The weight of the object on top of the beam (m2 * g) also acts downward.
Again, since the beam is in static equilibrium, the sum of these forces along the y-axis must be zero:
T * sin(θ) - m1 * g - m2 * g = 0 -----(2)
Now, we can solve these two equations to find the tension in the cable (T) and the horizontal force exerted by the wall (F).
(a) To find T, plug equation (1) into equation (2):
T * sin(θ) - m1 * g - m2 * g = T * cos(θ)
Rearranging the equation:
T * (sin(θ) - cos(θ)) = m1 * g + m2 * g
Dividing both sides by (sin(θ) - cos(θ)):
T = (m1 * g + m2 * g) / (sin(θ) - cos(θ))
Substitute the given values: m1 = 300.0 kg, m2 = 50.0 kg, g = 9.8 m⋅s^(-2), and θ = 40.0° into the equation to calculate T.
(b) To find the horizontal component of the force exerted by the wall (Fx), substitute the tension value (T) from part (a) into equation (1):
T * cos(θ) - Fx = 0
Simplifying the equation:
Fx = T * cos(θ)
Substitute the given values: T = (calculated value from part (a)), and θ = 40.0° into the equation to calculate Fx.
(c) To find the vertical component of the force exerted by the wall (Fy), substitute the tension value (T) from part (a) into equation (2):
T * sin(θ) - m1 * g - m2 * g = 0
Simplifying the equation:
Fy = m1 * g + m2 * g - T * sin(θ)
Substitute the given values: m1 = 300.0 kg, m2 = 50.0 kg, g = 9.8 m⋅s^(-2), and θ = 40.0° into the equation to calculate Fy.