A 5500 N weight is held suspended in equilibrium by two cables. Cable 1 applies a horizontal force to the right of the object and has a tension T1. Cable 2 applies a force upward and to the left at an angle of 40.0 to the negative x-axis and has a tension T2. What is tension T1?

Answer: T1 = 6554.6N

Explanation:

Given;

Weight w = 5500N

Cable 1 = T1 acting on the horizontal

Cable 2 = T2 acting 40° to the horizontal

To solve this kind of question we must take the horizontal and vertical component of force at equilibrium.

Taking the summation of forces on the horizontal (x axis)

T1 - T2 (cos40°) = 0 ... 1

Summation of forces on the y axis (vertical)

T2 (sin40°) - W = 0 ... 2

From equation 2

T2 (sin40) = W

T2 = W/sin40 ... 3

From equation 1

T1 = T2 (cos40)

Substituting T2 = W/sin40

T1 = (W/sin40) (cos40)

T1 = 5500N (cos40/sin40)

T1 = 6554.6N