A uniform horizontal plank AB of length 5.0\text{ m} and mass 40\text{ kg} rests on two sm… — JAMB UTME

A uniform horizontal plank $AB$ of length $5.0\text{ m}$ and mass $40\text{ kg}$ rests on two smooth supports, $C$ and $D$ , located $1.0\text{ m}$ from end $A$ and $1.0\text{ m}$ from end $B$ respectively. What is the maximum mass, in kilograms, of an object that can be placed at end $A$ without causing the plank to tilt?

Answer: 60 kg

Answer

The maximum mass that can be placed at end A without tilting the plank is 60 kg.

Just before tilting, the plank rotates around support C, causing the normal reaction at support D to drop to zero. Equating the anticlockwise moment of the added mass at end A about support C (

(m \cdot g) \times 1.0\text{ m}

) to the clockwise moment of the plank's weight about support C (

(40 \cdot g) \times 1.5\text{ m}

) yields

m = 60\text{ kg}

Step-by-Step Solution

Identify the tipping condition and pivot point

Support C acts as the pivot; the reaction at support D becomes zero (

R_D = 0

When extra weight is added at end A, the plank rotates about C and lifts off support D.

Calculate perpendicular distances from the pivot C

Distance to added mass

m

1.0\text{ m}

; Distance to plank's center of mass =

2.5\text{ m} - 1.0\text{ m} = 1.5\text{ m}

The weight of a uniform beam acts at its midpoint (2.5 m from either end).

Equate clockwise and counterclockwise moments about C

m \times g \times 1.0\text{ m} = 40\text{ kg} \times g \times 1.5\text{ m}

, giving

m = 60\text{ kg}

For the plank to remain balanced just before tipping, total anticlockwise moment must equal total clockwise moment.

Key Concept

Rotational Equilibrium and Tilting of Rigid Bodies

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic