Consider the 667 N weight held by two cables
shown below. The left-hand cable had tension
T2 and makes an angle of θ2 with the ceiling.
The right-hand cable had tension 540 N and
makes an angle of 55◦ with the ceiling.
The right-hand cable makes an angle of 55◦
with the ceiling and has a tension of 540 N.
a) What is the tension T2 in the left-hand cable slanted at an angle of θ2 with respect to the wall? Answer in units of N.
b) What is the angle θ2 which the left-hand cable makes with respect to the ceiling?
To find the answer to part a:
We have the weight of 667 N being held by two cables. The right-hand cable has a tension of 540 N.
The left-hand cable has tension T2 and makes an angle of θ2 with the ceiling.
To find the tension T2 in the left-hand cable, we can use the equilibrium of forces. Since the weight is being held in place, the sum of the vertical forces in equilibrium should be zero.
The vertical force on the weight is given by the weight itself, which is 667 N, and this force is balanced by the vertical components of the tension forces of both cables.
The vertical component of the tension in the right-hand cable is given by 540 N * sin(55°). This can be calculated as 540 N * sin(55°) ≈ 440.2 N.
The vertical component of the tension in the left-hand cable is given by the T2 * sin(θ2).
Now, setting up the equation:
667 N = T2 * sin(θ2) + 440.2 N
To solve for T2, we need the value of θ2.
Now, let's move on to part b:
We are given that the right-hand cable makes an angle of 55° with the ceiling.
To find the angle θ2 which the left-hand cable makes with respect to the ceiling, we need to use the right triangle formed by the tension force in the left-hand cable, the vertical component of the tension force in the left-hand cable, and the magnitude of the tension force in the left-hand cable.
Using the inverse sine function, we can solve for the angle θ2:
θ2 = arcsin((667 N - 440.2 N) / T2)
To find the values for T2 and θ2, we need to use the given information about the right-hand cable and solve the equations simultaneously.