A store owner wishes to hang a sign weighing 750 N so that Cable A attached to the store makes a 30 degree angle from the vertical wall. Cable B is attached horizontally to an adjoining building.
Calculate the necessaryt tension on Cable B.
To calculate the tension on Cable B, we first need to find the vertical and horizontal components of the force exerted by the sign.
Given:
Weight of the sign = 750 N
Angle made by Cable A with the vertical wall = 30 degrees
To find the vertical component of the force exerted by the sign, we can use the formula:
Vertical component = Weight * sin(angle)
Vertical component = 750 N * sin(30 degrees)
Vertical component = 750 N * 0.5
Vertical component = 375 N
To find the horizontal component of the force exerted by the sign, we can use the formula:
Horizontal component = Weight * cos(angle)
Horizontal component = 750 N * cos(30 degrees)
Horizontal component = 750 N * 0.866
Horizontal component = 649.5 N
Since Cable B is attached horizontally to an adjoining building, the tension in Cable B would be equal to the horizontal component of the force exerted by the sign.
Therefore, the necessary tension on Cable B is 649.5 N.
To calculate the necessary tension on Cable B, we can use trigonometric equations and the concept of equilibrium.
First, let's draw a free-body diagram of the sign:
^ Cable A
|
|
| /
| /
| /
| / Sign
| /
|/
-------------- Cable B
Now, let's break down the forces acting on the sign. The weight of the sign (750 N) acts vertically downwards (opposite to Cable A).
The tension in Cable A can be resolved into two components: one in the vertical direction and one in the horizontal direction. The vertical component of Cable A will balance out the weight of the sign, so the sum of the vertical forces is zero.
Using trigonometry, we can find that the vertical component of Cable A is given by:
Vertical component of Cable A = Tension in Cable A * cos(angle)
Since the sign is in equilibrium, the sum of the horizontal forces is zero as well. The only horizontal force acting on the sign is the tension in Cable B.
Now, to find the necessary tension on Cable B, we can set up the equation:
Tension in Cable B = Tension in Cable A * sin(angle)
Substituting the given values:
Angle = 30 degrees
Weight of the sign = 750 N
We need to find the tension in Cable B.
First, we need to find the tension in Cable A using the known weight of the sign:
Tension in Cable A = Weight of the sign / cos(angle)
Tension in Cable A = 750 N / cos(30 degrees)
Next, we can find the tension in Cable B:
Tension in Cable B = Tension in Cable A * sin(angle)
Tension in Cable B = (750 N / cos(30 degrees)) * sin(30 degrees)
After calculating these values, we will get the required tension in Cable B.