Q. A 200 kg load is suspended by a cable from a crane . Two

Question

ropes are attached to the load , acting at 90 deg to each
other. One rope exerts a pull of 220 N , at 10 deg below
horizontal, the other rope exerts a force of 360 N at 15 deg
below horizontal.
Calculate 1)tension in cable
2)angle of the cable to vertical.

Answer 1

Let's break the forces into their horizontal and vertical components.

Horizontal components:
Force 1: F1x = -220 * cos(10) = -216.08 N (negative because it's pointing left)
Force 2: F2x = 360 * cos(15) = 348.01 N

The net horizontal force (NH) is the sum of these two forces:
NH = F1x + F2x = -216.08 + 348.01 = 131.93 N

Vertical components:
Force 1: F1y = -220 * sin(10) = -38.15 N (negative because it's pointing downward)
Force 2: F2y = -360 * sin(15) = -93.44 N

The net vertical force (NV) is the sum of these forces:
NV = F1y + F2y = -38.15 - 93.44 = -131.59 N

Since the load is in equilibrium, the tension in the cable T must balance these forces. The vertical component of T, Fty must be equal to NV in magnitude but opposite in direction, and the horizontal component of T, Ftx must be equal to NH in magnitude but opposite in direction:
Fty = 131.59 N
Ftx = -131.93 N

To find the magnitude of T:
T = sqrt(Fty^2 + Ftx^2) = sqrt(131.59^2 + (-131.93)^2) = 186.20 N

Now let's find the angle θ of the cable to the vertical. We can use the tangent function:
tan(θ) = Ftx / Fty = -131.93 / 131.59

θ = arctan(-131.93 / 131.59)= -45.15 degrees (approx)

Since θ is negative, the angle is actually below the vertical axis in the opposite direction.

So the tension in the cable is 186.20 N and the angle of the cable to the vertical is approximately 45.15 degrees below vertical.