1

Question

Three equal masses of 1 kg each are placed at the vertices of an equilateral triangle PQR and a mass of 2 kg is placed at the centroid O of the triangle which is at a distance of √2 m from each of the vertices of the triangle. The net force (in Newton), acting on the mass of 2 kg is

Open in App

Solution

The correct option is **D** zero

Given:

OP=OQ=OR=√2 m

m=1 kg; M=2 kg

The gravitational force on mass 2 kg due to mass 1 kg at P.

FOP=G2×1(√2)2=G along OP

Similarly,

FOQ=G2×1(√2)2=G along OQ

FOR=G2×1(√2)2=G along OR

Net force at O in direction parallel to the base of triangle is due to force FOQ and FOR .

FOQcos30∘ and FORcos30∘ are equal and acting in opposite directions, thus balance each other.

Therefore net force at O will be in vertical direction.

The resultant force on the mass 2 kg at O is given by

Fnet=FOP−(FOQsin30∘+FORsin30∘)

⇒Fnet=G−(G2+G2)=0

Hence, option (d) is correct.

Given:

OP=OQ=OR=√2 m

m=1 kg; M=2 kg

The gravitational force on mass 2 kg due to mass 1 kg at P.

FOP=G2×1(√2)2=G along OP

Similarly,

FOQ=G2×1(√2)2=G along OQ

FOR=G2×1(√2)2=G along OR

Net force at O in direction parallel to the base of triangle is due to force FOQ and FOR .

FOQcos30∘ and FORcos30∘ are equal and acting in opposite directions, thus balance each other.

Therefore net force at O will be in vertical direction.

The resultant force on the mass 2 kg at O is given by

Fnet=FOP−(FOQsin30∘+FORsin30∘)

⇒Fnet=G−(G2+G2)=0

Hence, option (d) is correct.

Why this question: Key Concept: Principle of Superposition - The gravitational force between two masses is independent of other masses in the vicinity and the net force on a mass is equal to the vector sum of forces by all the masses. |

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program