Free body diagram Homework Problem

[Physically Impaired]

Can someone please help with this problem. I'm lost!

1. A 150 kg bear walks out onto a beam as illustrated in Fig P8.20 trying to get to a picnic
basket with mass 10kg. The beam is 6m long and has mass 40kg (you can consider its
weight to be concentrated at a point 3m from the wall.
(a) When the bear is 1m from the wall, how much tension will be in the wire? (The wire
is at an angle of 60o above the horizontal, directed toward the wall.)
(b) What is the horizontal component of the force of the wall on the beam at this point?
(c) What is the vertical component of the force of the wall on the beam at this point?
(d) If the wire can support a tension of 1500N, what is the maximum distance from the
wall the bear can walk.
(e) What is the moment of inertia of the bear at this point assuming its moment of inertia
is mx2, where m is the bear’s mass and x is the distance from the wall.
(f) What is the moment of inertia of the basket, again assuming its moment of inertia is
ml2 where m is the basket’s mass and l is the length of the beam.
(g) What is the moment of inertia of the beam assuming its moment of inertia is 1
3
ml2,
where m is the mass of the beam, and l is the length of the beam.
(h) What is the total moment of inertia (add the last three answers).
(i) If the wire was to snap with the bear at this point, what would be the angular acceleration
of the beam, bear and basket immediately after the wire snaps (assume they all
accelerate as one object with the moment of inertia as given above).
(j) What would be the bear’s tangential acceleration?

 

[CartoonKid]

This problem is quite direct. All you need to do is to draw a free body diagram. Remember that when dealing with static problems, FBD is essential to solve them.

Since you didn't mention about the location of the basket, I will just assume that it is at the end of the beam. Assign A as the contact point of the beam with the wall. Using \sum{M_A}=0, you can find out the tension of the rope. After getting the tension of the rope, the rest of the question can be solved easily.

 

[wais]

First, let's draw a free body diagram for the problem:

1. Bear:
- Weight (150kg) pointing downwards
- Tension in the wire (unknown) pointing upwards and to the left at a 60 degree angle
- Normal force from the beam (unknown) pointing upwards and to the right

2. Picnic basket:
- Weight (10kg) pointing downwards
- Tension in the wire (unknown) pointing upwards and to the left at a 60 degree angle

3. Beam:
- Weight (40kg) concentrated at a point 3m from the wall, pointing downwards
- Tension in the wire (unknown) pointing upwards and to the left at a 60 degree angle
- Normal force from the wall (unknown) pointing upwards and to the right

Now, let's solve for the unknowns in each part of the problem:

(a) Using the free body diagram for the bear, we can set up the following equation:
ΣFy = 0
N - 150kg(9.8m/s²)sin60° = 0
N = 735N
Therefore, the tension in the wire is 735N.

(b) Using the free body diagram for the beam, we can set up the following equation:
ΣFx = 0
Fx = 0
Therefore, the horizontal component of the force of the wall on the beam is 0N.

(c) Using the free body diagram for the beam, we can set up the following equation:
ΣFy = 0
N + Tsin60° - 40kg(9.8m/s²) = 0
N = 294N
Therefore, the vertical component of the force of the wall on the beam is 294N.

(d) To find the maximum distance from the wall the bear can walk, we can set up the following equation using the free body diagram for the bear:
ΣFx = 0
Tcos60° = 150kg(9.8m/s²)
T = 1470N
Therefore, the maximum distance from the wall the bear can walk is 1470N/(150kg(9.8m/s²)) = 1.02m.

(e) The moment of inertia of the bear can be calculated using the formula I = mx². Plugging in the values, we get:
I = (150kg)(