There is a right angle triangle:
-.-
-...-
-......-
-.........- T (wire)
-............-
-...............-
------------------ 30 degrees
....--.....--.....rod
....--.....--.....
....--.....--.....
....---------.....
....-.......-.....
....-.......-.....
....-.......-.....
....---------.....
sign 120n downwards
The weight of the sign and rod act through the centre point of the ro. The value w is 120n. The anle between wire and rod is 30 degrees.
Calculate the tension in the wire
The mark scheme says T=60/sin 30 or 60cos60
T=120N
But i dont understand where the numbers came from and how it was worked out please write down how you do it step by step!
Up force from wire at tip of rod = T sin 30
Down force at rod center = 120 N
moments about rod at wall
T sin 30 (L) = 120 (L/2) = 60 L
T = 60 /sin 30 = 60/0.5 = 120
the sign in the center of the rod? Let L be the length of the rod.
so it is critical to understand the rod connection to the wall. Is there any vertical force exerted on the wall, and is the wall exerting any moment on the rod.
without that knowledge, the mark scheme is useless.
To calculate 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.
1. First, let's analyze the forces acting on the rod. We have the weight of the rod acting vertically downwards, which is given as 120 N. This force can be represented as the vertical component of the tension in the wire.
2. Now, let's consider the forces acting on the wire. We have the tension in the wire acting horizontally, which is balancing the horizontal component of the weight of the rod. Since the angle between the wire and the rod is given as 30 degrees, we need to find the horizontal component of the 120 N weight.
3. To find the horizontal component, we can use trigonometry. The horizontal component is given by the formula:
Horizontal component = weight * cos(angle)
Plugging in the values:
Horizontal component = 120 N * cos(30 degrees)
4. Evaluating this expression:
Horizontal component = 120 N * 0.866 (cosine of 30 degrees is 0.866)
Horizontal component = 103.92 N
5. Since the tension in the wire is balancing this horizontal component, the tension in the wire is also equal to 103.92 N.
So, the tension in the wire is approximately 103.92 N. The mark scheme suggests an alternative solution using trigonometric ratios, where T = 60/sin(30) or T = 60*cos(60). Both of these expressions are equivalent and would yield the same result of approximately 103.92 N as well.