A traffic sign is hanging from two wires. If the weight of the sign is 35 N, what are the tensions in the two wires as shown below?
Traffic sign hanging by two wires, making an angle of 56 and 45
T1*sin(180-56) + T2*sin45 = 35 N,
Eq1: 0.83T1 + 0.71T2 = 35.
T1*Cos(180-56) + T2*Cos45 = 0,
Eq2: -0.56T1 + 0.71T2 = 0,
Subtract Eq2 from Eq1:
Diff.: 1.39T1 = 35,
T1 = 25.2 N.
In Eq1, replace T1 with 25.2 and solve for T2.
the sign isn't moving , so the horizontal components are equal and opposite
the vertical components sum to 35 N
"making an angle of 56 and 45" ... with what?
... a little more description would help
Well, I don't have a sign hanging from two wires, but I do know a thing or two about tension and angles. Let's give it a shot!
First, let's label the tensions in the wires as T1 and T2. Now, we can use some trigonometry to find the values.
In triangle ABC, where A is the angle between T1 and the vertical line and B is the angle between T2 and the vertical line, we have:
T1/sin(A) = T2/sin(B)
Since we know the angles, we can substitute the given values:
T1/sin(56) = T2/sin(45)
Now, we have one equation with two unknowns. But fear not, young mathematician! We also know that the sum of the vertical components of the tensions should equal the weight of the sign. So we can write another equation based on that:
T1*cos(56) + T2*cos(45) = 35 N
Now, we can solve these two equations simultaneously to find the tensions T1 and T2. I'm not quite skilled in math, but I'm sure you can handle it! Good luck!
To find the tensions in the two wires, we can use the concept of vector resolution.
Step 1: Draw a diagram of the problem.
Visualize the traffic sign hanging from the two wires, making angles of 56° and 45° with the horizontal direction. Label the unknown tensions in the wires as T1 and T2 respectively.
Step 2: Resolve the weight vector.
Split the weight vector into two components, one parallel to the wire T1 and the other parallel to the wire T2.
For the component parallel to T1, we use the sine function:
Weight component parallel to T1 = Weight * sin(angle between weight vector and T1)
For the component parallel to T2, we use the sine function:
Weight component parallel to T2 = Weight * sin(angle between weight vector and T2)
Step 3: Apply Newton's second law in the vertical direction.
Since the sign is in equilibrium (not accelerating vertically), the sum of the vertical forces must be zero. Considering the vertical direction, we have:
T1 * cos(56°) + T2 * cos(45°) = Weight
Step 4: Solve for the unknowns.
Using the equations from steps 2 and 3, we can substitute the weight components and solve for T1 and T2.
T1 * cos(56°) + T2 * cos(45°) = Weight
T1 * cos(56°) + T2 * cos(45°) = 35 N
Finally, solve the system of equations to find the unknown tensions T1 and T2.
Draw a position diagram showing the 2 wires and the suspended sign.
A very standard question if I assume the wires make angles of 45° and 56° along a horizontal.
draw a vector diagram of a triangle ABC, where AB is a vertical line representing the 35 N. AC the tension in the wire making a 45° angle, and CB the tension of the wire making the 56° angle. Simple geometry using parallel lines will given you
angle A = 45° , angle B = 34° and angle C = 101°
by the sine law:
BC/sin45 = 35/sin101 ----> BC = 35sin45/sin101 = appr 25.2 N
AC/sin34 = 34/sin101 ----> AC = 34sin45/sin101 = appr 24.5 N