---
title: "Find forces P and Q for equilibrium"
subject: "Applied Mechanics"
chapter: 2
key: "semester_1/applied_mechanics:unknown-forces-p-q-equilibrium"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/unknown-forces-p-q-equilibrium
site: BCT Engineering
---

# Find forces P and Q for equilibrium

Applied Mechanics · Chapter 2: Basic Concept in Statics and Static Equilibrium

- Asked in: 2074 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/unknown-forces-p-q-equilibrium
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2

## Question

Determine the values of the unknown forces P and Q for the system of forces to be in equilibrium. [Figure: concurrent forces: 30 kN acting to the left, Q acting to the right, 25 kN acting downward, and P acting up-left at 60° as marked.]

## Short answer

From ΣFy=0: P sin60° = 25 kN, so P = 28.87 kN. From ΣFx=0: Q = 30 + P cos60° = 44.43 kN.

## Answer

**Reading of the figure:** four concurrent forces: 30 kN to the left, Q to the right, 25 kN downward, and P acting up and to the left at 60° to the horizontal (the 60° is marked between P and the horizontal axis).

#### Equilibrium equations
ΣFy = 0 (up positive):
P sin 60° − 25 = 0
P = 25 / 0.8660 = 28.87 kN

ΣFx = 0 (right positive):
Q − 30 − P cos 60° = 0
Q = 30 + 28.87 × 0.5 = 30 + 14.43 = 44.43 kN

**Answer: P = 28.87 kN, Q = 44.43 kN.**

## Check yourself

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

1. Which equation is used first to find P?
   A. Any, as all contain P and Q
   B. SFy = 0, because only P and the 25 kN force have vertical components
   C. Moments about O
   D. SFx = 0, because Q is unknown
   Answer: B. Q and 30 kN are horizontal, so SFy involves only P. (Key point: Method)

2. P sin 60° = 25 gives P =
   A. 50.00 kN
   B. 28.87 kN
   C. 21.65 kN
   D. 14.43 kN
   Answer: B. P = 25/0.8660 = 28.87 kN. (Key point: Value of P)

3. Q - 30 - P cos 60° = 0 gives Q =
   A. 15.57 kN
   B. 44.43 kN
   C. 30 kN
   D. 58.87 kN
   Answer: B. Q = 30 + 28.87 x 0.5 = 44.43 kN. (Key point: Value of Q)

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