A drone flies at 10 m/s, 25° above the horizontal. Enter the horizontal component of its velocity in m/s, to one decimal place.
Gideon hauls a loaded wagon with a rope. The rope applies a 100 N force to the wagon at 30° above the horizontal.
Part of that force moves the wagon forward, and part of it lifts. How much of each?
You resolved velocity vectors into perpendicular components in the vectors topic. A force is a vector too, so the identical routine answers this.
A force at an angle is replaceable by two perpendicular components: a horizontal one and a vertical one.
The components come from the same two lines you used for velocities: F_x = F cos θ and F_y = F sin θ, with θ measured above the horizontal.
Nothing new is derived here. Only the quantity changed, so the components come out in newtons.
For Gideon's rope, with cos 30° = 0.87: F_x = 100 × 0.87 = 87. The horizontal component is 87 N.
And with sin 30° = 0.50: F_y = 100 × 0.50 = 50. The vertical component is 50 N.
So 87 N of the rope's force moves the wagon forward, and 50 N of it lifts.
A rope applies a 100 N force at 30° above the horizontal. Take cos 30° = 0.87 and sin 30° = 0.50. Calculate both components of the force.
Write down the values in the question:
F = 100 N
θ = 30°
F_x = ?
F_y = ?
Write down the equations:
F_x = F cos θ
F_y = F sin θ
Substitute in the values:
F_x = 100 × 0.87
F_y = 100 × 0.50
Simplify:
F_x = 87 N
F_y = 50 N
The horizontal component is 87 N, and the vertical component is 50 N.
A cable applies a 40 N force at 53° above the horizontal. Take cos 53° = 0.60 and sin 53° = 0.80. Calculate both components of the force.
Write down the values in the question:
F = 40 N
θ = 53°
F_x = ?
F_y = ?
Write down the equations:
F_x = F cos θ
F_y = F sin θ
Substitute in the values:
F_x = 40 × 0.60
F_y = 40 × 0.80
Simplify:
F_x = 24 N
F_y = 32 N
The horizontal component is 24 N, and the vertical component is 32 N.
A kite line applies a 55 N force at 37° above the horizontal. Take sin 37° = 0.60 and cos 37° = 0.80. Enter the vertical component in N.
A tow strap applies an 80 N force at 53° above the horizontal. Take sin 53° = 0.80 and cos 53° = 0.60. Enter the horizontal component in N.
A 90 N force acts at 37° above the horizontal. Which calculation gives its vertical component?
A leash applies a 120 N force at 30° above the horizontal. Take sin 30° = 0.50 and cos 30° = 0.87. Enter the vertical component in N.
A winch cable applies a 140 N force at 53° above the horizontal. Take sin 53° = 0.80 and cos 53° = 0.60. Enter the vertical component in N.
June drags a sled by a rope that applies a 65 N force at 37° above the horizontal. Take sin 37° = 0.60 and cos 37° = 0.80. Enter the component that moves the sled forward, in N.