A hungry bear weighs 700 N walks to a basket of candy that is located in
end of a beam . The beam is uniform and weighs 200N , with a length of 6.0 m . The beam is supported
by a cord forming an angle of 60 ° . The basket weighs 80 N. When the bear is within 1.2 m
determine the tension of the rope.
I know the bear and the bear are affected by the mass- gravity force and the normal, the cord is affected by the tension and the mass- gravity force, but i really don't know how to start it.
any help will be appreciated
the two conditions of equilibrium are:
Sum of vertical and horizontal forces is zero.
Sum of moments about any point is zero.
Without knowing how the rope is attached, I can't do more. But here are some similar moment related problems that might help you understand.
Try this one: https://www.youtube.com/watch?v=EzLnvCJxW54
To determine the tension in the rope, we need to analyze the forces acting on the system.
First, let's consider the forces acting on the bear:
1. Weight (or gravitational force) - The bear weighs 700 N, so there is a downward force acting on it.
Next, let's consider the forces acting on the beam:
1. Weight (or gravitational force) - The beam weighs 200 N, so there is a downward force acting on it.
2. Tension - The tension in the rope is pulling the beam upward.
Finally, let's consider the forces acting on the basket of candy:
1. Weight (or gravitational force) - The basket weighs 80 N, so there is a downward force acting on it.
Now, let's break down the forces acting along the beam:
1. Vertical forces - The sum of the vertical forces acting on the beam must be zero since it is in equilibrium.
∑Fy = T - 700 N - 200 N - 80 N = 0
Simplifying the equation, we get:
T - 980 N = 0
Therefore, the tension in the rope is equal to 980 N.