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1) Represent each force in x and y components.
using the provided x and y axes as positive:
*note: make sure your calculator is in degrees and not radians
F1x = 30cos45° = 21.21 lb F1y = 30sin45° = 21.21 lb
F2x = 40cos15° = 38.64 lb F2y = 40sin15° = -10.35 lb
F3x = 25sin15° = 6.47 lb F3y = -25cos15° = -24.15 lb
2) Determine the force in each cable. The mass of the crate is 200 kg. Cord BC remains horizontal, and AB has a length of 1.5m. Use y=0.75m.
From geometry, we can find the angle AB makes with the horizontal:
theta = sin-1(0.75/1.5) = 30°
Apply the equations of equilibrium:
sum F's in the x direction: BAcos30° - BC = 0 [1]
sum F's in the y direction: BAsin30° - mg = 0 [2]
BA = mg/sin30° = (200)(9.81)/sin30° = 3.92 kN
subbing into [1]: BC = 3.40 kN
Mark Yourself Question
- Grab a piece of paper and try this problem yourself.
- When you're done, check the "I have answered this question" box below.
- View the solution and report whether you got it right or wrong.
1) Represent the 800 N force in x and y-components:
2) Determine the components of the force along lines AC and BC:
3) Determine the resultant force along lines AC and BC:
practice quiz: Vector Operations
Find the components of F along the u and v axis. Take F = 100 N.
Find the components of the 450lb force along AC and AB.
Determine the angles between the forces and the flagpole.
