---
title: "Force P to accelerate 600 N block"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:block-600n-force-p-acceleration-dynamic-equilibrium"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-600n-force-p-acceleration-dynamic-equilibrium
site: BCT Engineering
---

# Force P to accelerate 600 N block

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2073 Chaitra, 8+2 marks
- Video: [D Alemberts Principle ||Problem 2 ||Kinetics of Particles|| Engineering Mechanics](https://www.youtube.com/watch?v=m-xKXRFh_IY) — Engineering Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/block-600n-force-p-acceleration-dynamic-equilibrium
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

A 600 N block rests on a horizontal plane. Find the magnitude of P required to produce the block an acceleration of 2 m/s² to the right. The coefficient of friction is 0.25. What do you mean by dynamic equilibrium? [Figure: block of 600 N pulled by force P acting at 20° above the horizontal.]

## Short answer

Vertical: N = 600 – 0.342P. Horizontal: 0.9397P – 0.25(600 – 0.342P) = (600/9.81)×2, giving P = 265.6 N. Dynamic equilibrium: include an inertia force ma opposite acceleration, then ΣF = 0.

## Answer

#### Reading of the figure
The 600 N block is on a horizontal plane. P pulls the block at 20° above the horizontal, so its vertical component P sin 20° lifts the block slightly and the normal reaction is N = W − P sin 20°. μ = 0.25 and a = 2 m/s² to the right.

```text
          P
         /
        / 20°
   ____/
  |600N|--> a = 2 m/s²
 ------------------
```

#### Equations of motion
- Vertical: N = 600 − P sin 20° = 600 − 0.342P
- Horizontal: P cos 20° − μN = (W/g) a
 P(0.9397) − 0.25(600 − 0.342P) = (600/9.81)(2)
 0.9397P − 150 + 0.0855P = 122.32
 1.0252P = 272.32

**P = 265.6 N**

(Check: N = 600 − 0.342(265.6) = 509.2 N, friction = 0.25 × 509.2 = 127.3 N, and 265.6 cos 20° − 127.3 = 122.3 N = (600/9.81)(2) ✓.)

#### Dynamic equilibrium
A body accelerating under a resultant force can be treated as if it were in equilibrium (D'Alembert's principle) by adding an imaginary **inertia force** equal to m a, directed opposite to the acceleration:
ΣF − m a = 0
Here the inertia force = (600/9.81)(2) = 122.3 N acts to the left on the block, and the usual equations of statics (ΣFx = 0, ΣFy = 0, ΣM = 0) can then be applied. The block is then said to be in *dynamic equilibrium*.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With P pushing down at 20 degrees on the block, what is the normal reaction?
   A. N = 600
   B. N = 600 - P sin 20
   C. N = 600 + P sin 20
   D. N = 600 + P cos 20
   Answer: B. P pushes downward with component P sin 20, so N = W + P sin 20, which increases friction. (Key point: Normal reaction with inclined P)

2. Which horizontal equation of motion is used?
   A. P cos 20 = mu N
   B. P cos 20 - mu N = (600/9.81)(2)
   C. P sin 20 - mu N = m a
   D. P - mu N = 0
   Answer: B. The net horizontal force P cos 20 - mu N equals m a, with m = 600/9.81 and a = 2 m/s^2. (Key point: Horizontal equation)

3. Solving 0.9397P - 0.25(600 + 0.342P) = 122.32 gives which P?
   A. 265.6 N
   B. 122.3 N
   C. 150 N
   D. 318.8 N
   Answer: A. 0.8542P = 272.32, so P = 318.8 N. The value 265.6 N applies only if P pulled upward. (Key point: Result for P)

4. What is dynamic equilibrium (D'Alembert's principle)?
   A. Adding an inertia force m a opposite to the acceleration so the equations of statics can be applied
   B. A body moving at constant velocity
   C. A body at rest with zero friction
   D. Balance of kinetic and potential energy
   Answer: A. Adding the inertia force -m a gives sum F - m a = 0, so a moving body can be solved with statics. (Key point: Dynamic equilibrium)

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