A 195.0-N sign is suspended from a horizontal strut of negligible weight. The force exerted on the strut by the wall is horizontal. Draw an FBD to show the forces acting on the strut. (Do this on paper. Your instructor may ask you to turn in this work.)The angle is 30 degree
Find the tension T in the diagonal cable supporting the strut.
N
I tried 195sin(30)= 97.5 N, but its wrong. Please someone help me out. Thank you. Sorry for re-posting.
tension*sin30=195N
ΣFx: -Tsin30° + 200N = 0
The force (Tsin30) is acting in the opposite direction to the 200N. This has to be true in order for them to add to zero. If the forces are acting in opposite directions then they must have equal magnitudes for it to stay in place.
T = 400 N
To find the tension T in the diagonal cable supporting the strut, we can start by drawing a free body diagram (FBD) to visualize the forces acting on the strut.
Here's how you can draw the FBD on paper:
1. Draw a horizontal line to represent the strut.
2. At one end of the line, draw a vertical line to represent the weight of the sign (195.0 N) acting downward.
3. At the other end of the horizontal line, draw a diagonal line at a 30-degree angle upward to represent the tension in the diagonal cable (T). Remember to label the angle as 30 degrees.
4. Finally, draw a horizontal line towards the left to represent the force exerted by the wall on the strut.
Now, to find the tension T, we will use the concept of equilibrium. In equilibrium, the sum of the forces in both the horizontal and vertical directions is zero.
In the horizontal direction, there is only one force, which is the force exerted by the wall. Let's call this force F_wall.
In the vertical direction, there are two forces: the weight of the sign acting downward and the tension in the diagonal cable acting upward. Let's call the weight of the sign F_weight.
Using trigonometry, we can break down the weight of the sign into its vertical and horizontal components. The vertical component of the weight is given by F_weight_vertical = F_weight * sin(theta), where theta is the angle of 30 degrees.
Now, since the strut is in equilibrium, we have:
Sum of horizontal forces = 0
F_wall = 0
Sum of vertical forces = 0
F_weight_vertical + T = 0
F_weight * sin(theta) + T = 0
Since we know the weight of the sign is 195.0 N, we can substitute this value into the equation:
195.0 * sin(30) + T = 0
Now, let's solve for T:
T = -195.0 * sin(30)
Calculating this expression, we get:
T = -97.5 N
Note that the negative sign indicates that the tension in the diagonal cable is acting in the opposite direction to the weight of the sign. However, in most cases, tension is represented as a positive value. Therefore, we can take the magnitude of T by removing the negative sign, giving us:
T = 97.5 N
So, the tension in the diagonal cable supporting the strut is 97.5 N.