A hungry bear weighing 720 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam (see the figure below). The beam is uniform, weighs 200 N, and is 7.00 m long, and it is supported by a wire at an angle of θ = 60.0°. The basket weighs 80.0 N.
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To find the tension in the wire supporting the beam, we need to consider the forces acting on the beam.
1. Weight of the bear: The bear weighs 720 N and acts downward at the center of mass of the bear.
2. Weight of the beam: The beam weighs 200 N and acts downward at its center of mass, which is also the point of support.
3. Weight of the basket: The basket weighs 80 N and acts downward at the end of the beam.
4. Tension in the wire: The tension in the wire acts upward at an angle of θ = 60.0°.
Let's break down the force components acting in the vertical and horizontal directions:
Vertical forces:
- The vertical component of the tension force in the wire will balance the sum of the vertical forces.
Horizontal forces:
- The horizontal component of the tension force in the wire will balance the horizontal forces on the beam.
Now, let's calculate the vertical forces:
- The weight of the bear acts downward with a force of 720 N.
- The weight of the beam acts downward with a force of 200 N.
- The weight of the basket acts downward with a force of 80 N.
Summing up the vertical forces:
Fy = 720 N + 200 N + 80 N
Next, let's calculate the horizontal forces:
- The horizontal component of the tension force in the wire will balance the horizontal forces on the beam.
Summing up the horizontal forces:
Fx = 0 (since there are no horizontal forces on the beam)
Now we can calculate the vertical component of the tension force:
Tension * cos(θ) = Fy
Solving for the tension:
Tension = Fy / cos(θ)
Substituting the values:
Tension = (720 N + 200 N + 80 N) / cos(60.0°)