- #1
davedave
- 50
- 0
A special fastener is used to anchor three cables to an east-facing wall in a factory. One cable applies a load of 300N straight down. The second cable applies a load of 400N, horizontally toward the south. The third cable applies a load of 500N toward the north, but angled at 30 degrees from the horizontal , down toward the floor. What is the total load on the fastener?
my solution
the vertical component of the 500N force = 500*sin30 = 250 N down
the north component of the 500 N force = 500*cos30 north
the sum of the north-south components = 300 north + 500*cos30 north + 400 south
Net north-south component = 300 + 500*cos30 - 400 = 333 N
The 333 N force vector and the 250 N force vector are perpendicular to each other. The vector that represents the sum of these 2 vectors is the hypotenuse of the right triangle formed by the 3 vectors.
Magnitude of total force = (250^2 + 333^2)^0.5 = 416 N
my solution
the vertical component of the 500N force = 500*sin30 = 250 N down
the north component of the 500 N force = 500*cos30 north
the sum of the north-south components = 300 north + 500*cos30 north + 400 south
Net north-south component = 300 + 500*cos30 - 400 = 333 N
The 333 N force vector and the 250 N force vector are perpendicular to each other. The vector that represents the sum of these 2 vectors is the hypotenuse of the right triangle formed by the 3 vectors.
Magnitude of total force = (250^2 + 333^2)^0.5 = 416 N
