PHYSICS · FORCES AND MOTION

What changes when you pull at an angle?

A 2 kg block rests on a horizontal surface. A 10 N force pulls it at an angle θ above the horizontal. Use g = 10 m/s².

Think about it

As θ increases, which force changes even though the block does not move vertically?

Show explanation

The upward component of the pull helps support the block, so the surface needs to push upward less. The normal force decreases.

Weight 20 N downward. Normal 15.00 N upward. Applied force 10 N at 30 degrees above the rightward horizontal. Horizontal component 8.66 N right; vertical component 5.00 N up. Friction 0.00 N and absent. Acceleration 4.33 metres per second squared right. Solid force tails meet at the centre; dashed components form the applied force triangle, not extra forces.surfaceN = 15.00 N ↑mg = 20 N ↓F = 10 N30°Fx = 8.66 N →Fy = 5.00 N ↑
Solid arrows are external forces. Dashed arrows resolve F and must not be counted again. Every vector uses the same scale: 6 diagram units per newton. Kinetic friction is disabled: fₖ = 0 N.
Fx
≈ 8.66 N
Fy
≈ 5.00 N
N
≈ 15.00 N
fk
≈ 0.00 N
ax
≈ 4.33 m/s²

Step by step

  1. Resolve the pull.Fₓ = F cos θ and Fᵧ = F sin θ.
  2. Use vertical equilibrium.While contact remains, vertical acceleration is zero: N + F sin θ − mg = 0. Therefore N = mg − F sin θ.
  3. Use the horizontal resultant.Without friction, ma = F cos θ, so a = F cos θ / m.
At θ = 30°:
Fₓ = 10 cos 30° ≈ 8.66 N
Fᵧ = 10 sin 30° = 5 N
N = 20 − 5 = 15 N
a = 8.66 / 2 ≈ 4.33 m/s²

What changes with friction?

Assume the block is already sliding to the right. Then fₖ = μₖN and aₓ = [F cos θ − μₖ(mg − F sin θ)] / m. At 30°, fₖ = 0.20 × 15 = 3 N and aₓ = (8.66 − 3) / 2 ≈ 2.83 m/s².

A negative acceleration means the right moving block is slowing down. This model does not continue through stopping and reversal without checking static friction.

Common misconception

N is not always mg. Here another force has an upward component, so the surface supplies less support.

Unit check

Force divided by mass gives N/kg = m/s², the correct unit for acceleration.

Check at 0° and 30°

At 0°: N = 20 N and a = 5 m/s². At 30°: N = 15 N and a ≈ 4.33 m/s².

Try it yourself

1. Find N and a without friction at 60°.

Fₓ = 5 N and Fᵧ = 8.66 N. So N = 11.34 N and a = 5/2 = 2.50 m/s².

2. At 0° with μₖ = 0.20, find the acceleration.

N = 20 N, fₖ = 4 N, and a = (10 − 4)/2 = 3.00 m/s².

Foundation → Core practice → Advanced extension

Build on this diagram

Begin with P1 and P2, then use P3 to distinguish static friction from its maximum and P5 to check loss of contact.

Contact and energy: the vertical loop in P8