A 98 N toolbox sits on a 30° ramp up to the bed of a truck.
Its weight points straight down. But the ramp's useful directions are tilted: parallel to the ramp face, and perpendicular to it.
Resolving the weight into those two directions is the same component routine you just ran, aimed at the ramp's own directions.
On an incline, resolve the weight into a component parallel to the ramp face and a component perpendicular to it.
The parallel component is w_par = mg sin θ. It is the part of the weight that tries to slide the object down the slope.
The perpendicular component is w_perp = mg cos θ. It is the part of the weight that presses the object into the ramp.
The ramp angle θ is the angle the ramp face makes with the horizontal.
For the toolbox, with sin 30° = 0.50: w_par = 98 × 0.50 = 49. So 49 N of the weight acts parallel to the ramp face, down the slope.
With cos 30° = 0.87: w_perp = 98 × 0.87 = 85.26, which is about 85 N pressing into the ramp.
A check that catches most slips: sin goes with the slide direction, and cos goes with the press direction. A steeper ramp grows the slide part and shrinks the press part.
A 98 N toolbox sits on a 30° ramp. Take sin 30° = 0.50 and cos 30° = 0.87. Calculate both components of its weight.
Write down the values in the question:
w = mg = 98 N
θ = 30°
w_par = ?
w_perp = ?
Write down the equations:
w_par = mg sin θ
w_perp = mg cos θ
Substitute in the values:
w_par = 98 × 0.50
w_perp = 98 × 0.87
Simplify:
w_par = 49 N
w_perp = 85.26 N
w_par = 49 N down the slope, and w_perp = 85.26 N, about 85 N, pressing into the ramp.
An 80 N crate rests on a 30° ramp. Take sin 30° = 0.50 and cos 30° = 0.87. Enter the weight component parallel to the ramp face, in N.
A 60 N planter rests on a 37° ramp. Take sin 37° = 0.60 and cos 37° = 0.80. Enter the weight component perpendicular to the ramp face, in N.
A 110 N paving stone rests on a 37° ramp. Take sin 37° = 0.60 and cos 37° = 0.80. Calculate both components of its weight.
Write down the values in the question:
w = mg = 110 N
θ = 37°
w_par = ?
w_perp = ?
Write down the equations:
w_par = mg sin θ
w_perp = mg cos θ
Substitute in the values:
w_par = 110 × 0.60
w_perp = 110 × 0.80
Simplify:
w_par = 66 N
w_perp = 88 N
w_par = 66 N down the slope, and w_perp = 88 N pressing into the ramp.
A 70 N bag sits on a 53° ramp. Which calculation gives the weight component parallel to the ramp face?
A 150 N keg rests on a 53° ramp. Take sin 53° = 0.80 and cos 53° = 0.60. Enter the weight component parallel to the ramp face, in N.
A 90 N toolbox rests on a 30° ramp. Take sin 30° = 0.50 and cos 30° = 0.87. Enter the weight component perpendicular to the ramp face, in N.
A 14 kg mower sits on a 37° ramp. Take g = 9.8 N/kg, sin 37° = 0.60 and cos 37° = 0.80. Enter the weight component parallel to the ramp face, in N.