A 1320-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the figure, find (a) the magnitude of the tension in the wire and the magnitudes of the (b) horizontal and (c) vertical components of the force that the wall exerts on the left end of the beam.
From the wall to the beam creates a 50 degree angle. From the beam to the crate is a 30 degree angle.
The drawing shows the beam and the five forces that act on it: the horizontal and vertical components S(x) and S(y) that the wall exerts on the left end of the beam, the weight W(b) of the beam, the force due to the weight Wc of the crate,
and the tension T in the cable. The beam is uniform, so its center of gravity is at the center of the beam, which is where its weight can be assumed to act. Since the beam is in equilibrium, the sum of the torques about any axis of rotation must be
zero ( Σ τ = 0) , and the sum of the forces in the horizontal and vertical directions
must be zero ( Σ F(x)= 0, Σ F(y)= 0) . These three conditions will allow us to determine the magnitudes of S(x), S(y), and T.
We will begin by taking the axis of rotation to be at the left end of the beam. Then
the torques produced by S(x) and S(y) are zero, since their lever arms are zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, T, in it. Setting the sum of the torques produced by the three forces equal to zero gives (with L equal to the length of the beam)
Σ τ = − W(b){0.5• L•cos30°} − W(c) •{Lcos30°} + T• {L•sin80°} = 0.
Algebraically eliminating L from this equation and solving for T gives
T = [W(b){0.5• L•cos30°} + W(c) •{Lcos30°}]/sin80º =
={1320• 0.5 •cos30°+1960• cos30°}/sin 80 º=2883 N.
Since the beam is in equilibrium, the sum of the forces in the vertical direction
must be zero:
Σ F(y) = + S(y) − W(b) − W(c) + T•sin50° =0
Solving for S(y) gives
S(y) =W(b)+W(c)-T•sin50 =1320+1960-2883•sin50°=1071 N.
The sum of the forces in the horizontal direction must also be zero:
Σ F(x)= + S(x) − T• cos50° = 0.
so that
S(x) = T• cos50° =2883•cos50° =1853 N.
Well, that's quite a tangled situation! But don't worry, I'm here to help you figure things out with a touch of humor.
First things first, let's name our forces so that we can keep track of them. We have the tension in the wire, which we're going to call "T", and the force that the wall exerts on the left end of the beam, which we'll call "Fw".
Now, let's break down the problem step by step. We'll start by finding the tension in the wire. To do that, we need to resolve the weight of the crate into horizontal and vertical components.
The weight of the crate is 1960 N, which we can split into two components: one parallel to the beam (horizontal) and the other perpendicular to it (vertical).
Using some trigonometry magic, we can find that the horizontal component is given by: Fh = 1960 N * cos(30°), and the vertical component is given by: Fv = 1960 N * sin(30°).
Now, let's move on to the forces acting on the beam. We have the weight of the beam itself, which we'll call "Wb", and the force that the wall exerts on the left end of the beam, "Fw". These forces must balance each other out to keep the beam in equilibrium.
Taking a comedic approach, we can say that the beam is trying really hard not to fall down and do a "free-fall" comedy routine. So the vertical component of the force, Fv, must balance the weight of the beam, Wb, for our comedic beam to remain stable.
Therefore, we can say that Fv = Wb, which gives us 1960 N * sin(30°) = Wb.
Now, to find the force that the wall exerts on the left end of the beam, Fw, we need to consider the horizontal component of the force acting on the beam. Since there are no other horizontal forces acting on the beam, Fh must balance out Fw.
So, Fh = Fw, which gives us Fh = 1960 N * cos(30°).
Finally, to get the tension in the wire, T, we need to balance the vertical forces acting on the beam. We can say that Fv + T = Wb.
So, T = Wb - Fv.
Now that we've cracked a few jokes, it's time for some serious calculations. Plug in the values you have and do the math to find the magnitude of the tension in the wire (T), and the magnitudes of the horizontal and vertical components of the force that the wall exerts on the left end of the beam (Fw).
Remember, sometimes laughter is the best way to approach a problem, even if it's about beams and crates!
To solve this problem, we can break down the forces into their components and then apply Newton's laws of motion.
First, let's break down the forces on the beam:
1. Resolve the weight of the beam into its horizontal and vertical components:
- The weight of the beam is given as 1320 N.
- The vertical component of the beam's weight is 1320 N * cos(50°).
- The horizontal component of the beam's weight is 1320 N * sin(50°).
2. Resolve the weight of the crate into its horizontal and vertical components:
- The weight of the crate is given as 1960 N.
- The vertical component of the crate's weight is 1960 N * cos(30°).
- The horizontal component of the crate's weight is 1960 N * sin(30°).
Now, let's apply Newton's laws of motion to find the tensions in the wire and the components of the force exerted by the wall:
(a) Tension in the wire:
- The total vertical forces acting on the beam are the vertical component of the beam's weight and the vertical component of the crate's weight.
- Tension in the wire is equal to the total vertical force acting on the beam.
- Tension in the wire = (1320 N * cos(50°)) + (1960 N * cos(30°)).
(b) Horizontal component of the force exerted by the wall:
- Since the beam is in equilibrium, the sum of the horizontal forces acting on the beam is zero.
- The only horizontal force is the horizontal component of the beam's weight.
- The horizontal component of the force exerted by the wall on the beam is equal in magnitude but opposite in direction to the horizontal component of the beam's weight.
- Horizontal component of the force exerted by the wall = -(1320 N * sin(50°)).
(c) Vertical component of the force exerted by the wall:
- Since the beam is in equilibrium, the sum of the vertical forces acting on the beam is zero.
- The vertical forces are the vertical component of the beam's weight, the vertical component of the crate's weight, and the tension in the wire (upward force).
- The vertical component of the force exerted by the wall is equal to the sum of the vertical components of the beam's weight and the crate's weight, minus the tension in the wire.
- Vertical component of the force exerted by the wall = (1320 N * cos(50°)) + (1960 N * cos(30°)) - (Tension in the wire).
Now you can substitute the given values and calculate the required quantities.
To find the magnitude of the tension in the wire, we can use the concept of equilibrium. In equilibrium, the sum of all the forces acting on an object is equal to zero.
Step 1: Resolve the forces into their horizontal and vertical components:
The weight of the crate (1960 N) can be resolved into its horizontal and vertical components.
Vertical component = 1960 N * sin(30°)
Horizontal component = 1960 N * cos(30°)
Step 2: Find the vertical and horizontal components of the force that the wall exerts on the left end of the beam:
The force that the wall exerts on the beam can be resolved into its vertical and horizontal components.
Vertical component = 1320 N * sin(50°)
Horizontal component = 1320 N * cos(50°)
Step 3: Apply the equilibrium condition:
The sum of the vertical components of the forces must be zero.
Vertical component of wall force + vertical component of tension - vertical component of crate weight = 0
Vertical component of wall force = -vertical component of tension + vertical component of crate weight
Substituting the values from step 2 and the weight component from step 1:
1320 N * sin(50°) = -Tension * sin(30°) + 1960 N * sin(30°)
Solving for the Tension, we can rearrange the equation:
Tension = (1320 N * sin(50°) - 1960 N * sin(30°)) / sin(30°)
Now, substitute the values given and calculate the magnitude of the tension.
For the horizontal component of the wall force, it will be equal to the horizontal component of the crate weight:
Horizontal component of wall force = Horizontal component of crate weight = 1960 N * cos(30°)
For the vertical component of the wall force, we have already found it in step 2.
Therefore:
(a) Magnitude of the tension in the wire = (1320 N * sin(50°) - 1960 N * sin(30°)) / sin(30°)
(b) Magnitude of horizontal component of force that the wall exerts on the left end of the beam = 1960 N * cos(30°)
(c) Magnitude of vertical component of force that the wall exerts on the left end of the beam = 1320 N * sin(50°)