Joe wishes to hang a sign weighing 730 N so that cable A, attached to the store, makes a 30.0° angle, as shown below. Cable B is horizontal and attached to an adjoining building. What is the tension in cable B?
thank you and the answer is 433 to anyone else trying to figure this out
To find the tension in cable B, we can use the concept of force equilibrium. In a state of equilibrium, the sum of the forces acting on an object is zero.
In this case, we have two forces acting on the sign: the weight of the sign (730 N) acting vertically downward and the tension in cable B acting horizontally.
First, let's resolve the weight of the sign into its vertical and horizontal components. The vertical component would be the weight multiplied by the sine of the angle:
Vertical component = Weight * sin(angle)
= 730 N * sin(30°)
Substituting the values:
Vertical component = 730 N * 0.5
= 365 N
The horizontal component of the weight is equal to the tension in cable B:
Horizontal component = Tension in cable B
Since the sign is in equilibrium, the sum of the horizontal forces should be zero:
Tension in cable B + Horizontal component = 0
Now, substituting the values:
Tension in cable B + 365 N = 0
To find the tension in cable B, we can rearrange the equation:
Tension in cable B = -365 N
The negative sign indicates that the tension in cable B is in the opposite direction of the horizontal component of the weight. Therefore, the tension in cable B is 365 N.
T sin 30 = force up = 730
T = 1460 N in cable A
T cos 30 = horizontal component = 1460 cos 30 = 1264 N