A particle of `m=5kg` is momentarily at rest at `x= at `t=. It is acted upon by two forces `vec(F)_(1)` and `vec(F)_(2)`. `Vec(F)_(1)=70hat(j)` N. The direction and manitude of `vec(F)_(2)` are unknown. The particle experiences a constant acceleration, `vec(a)`,in the direction as shown in (figure) Neglect gravity.


a.Find the missing force `vec(F)_(2)`.
b. What is the velocity vector of the particle at `t=10 s`?
c. What third force, `vec(F)_(3)` is required to make the acceleration of the particle zero? Either give magnitude and direction of `vec(F)_(3)` or its components.

Answer

a.`30sqrt(2)`; b.`60hat(i)+80hat(j)`; c.`30hat(i)-40hat(j)`

Explanation

`vec(F)_(1)+vec(F)_(2)=m.vec(a)`
`(70hat(j)+F_(2)hat(i)+F_(2)hat(j))=5(10 xcos53hat(i)+10sin53hat(j))`
`=50(10xx3/5hat(i)+10xx4/2hat(j))`
By comparing two sides. We get
`F_(2)hat(i)=30hat(i)rArrF_(2)hat(j)=-30hat(j)`
`:. vec(F)_(2)=30hat(i)-30hat(j)`
`|F_(2)| =sqrt((30)^(2)+(30)^(2))=30sqrt(2)N`
At `t`=`10s`
`v=vec(u)+at=10(6hat(i)+8hat(j))=60hat(i)+80hat(j)`
Acceleration will be zero if a third force makes resultant force =0
`:. vec(F)+vec(F)_(2)+vec(F)+(3)=0`
or `vec(F)_(3)= -(vec(F)_(1)+vec(F)_(2))=-(70hat(j)+30hat(i)-30hat(j))=30hat(i)-40hat(j)`