Question
Three forces start acting simultaneously on a particle moving with velocity, $$\vec v.$$ These forces are represented in magnitude and direction by the three sides of a triangle $$ABC.$$ The particle will now move with velocity
Three forces start acting simultaneously on a particle moving with velocity, $$\vec v.$$ These forces are represented in magnitude and direction by the three sides of a triangle $$ABC.$$ The particle will now move with velocity
A.
less than $${\vec v}$$
B.
greater than $${\vec v}$$
C.
$$\left| v \right|$$ in the direction of the largest force BC
D.
$${\vec v,}$$ remaining unchanged
Answer :
greater than $${\vec v}$$
Solution :
As shown in the figure, the three forces are represented by the sides of a triangle taken in the same order.
Therefore the resultant force is zero. $${{\vec F}_{net}} = m\vec a.$$
Therefore acceleration is also zero ie velocity remains unchanged.
As shown in the figure, the three forces are represented by the sides of a triangle taken in the same order.
Therefore the resultant force is zero. $${{\vec F}_{net}} = m\vec a.$$
Therefore acceleration is also zero ie velocity remains unchanged.
