A √34 m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If F_f and…

Question

A √34 m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_wF_f will be : (Use g=10 m s^(-2).)

MARKS Content Team · Accepted Answer

Correct answer: (C) 3√109. Applying force balance in vertical direction N_1=mg=100 N Applying torque balance at the bottom contact point, mg×3/2=N_2×5⇒N_2=F_w=30 N Applying force balance in horizontal direction f_s=N_2=30 N Net reaction at the floor will be, F_f=√(N1^2+fs^2)=√(100^2+30^2)=10√109 N Therefore, F_wF_f=30/(10√109)=3√109

A $\sqrt{34} m$ long ladder weighing $10 kg$ leans on a frictionless wall. Its feet rest on the floor $3 m$ away from the wall as shown in the figure. If $F_{f}$ and $F_{w}$ are the reaction forces of the floor and the wall, then ratio of $\frac{F_{w}}{F_{f}}$ will be :
(Use $g = 10 m s^{- 2}$ .)

Solution

Relevant Questions

A 34 m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If Ff and Fw are the reaction forces of the floor and the wall, then ratio of FwFf will be :(Use g=10 m s-2.)

Solution

Relevant Questions

A $\sqrt{34} m$ long ladder weighing $10 kg$ leans on a frictionless wall. Its feet rest on the floor $3 m$ away from the wall as shown in the figure. If $F_{f}$ and $F_{w}$ are the reaction forces of the floor and the wall, then ratio of $\frac{F_{w}}{F_{f}}$ will be :
(Use $g = 10 m s^{- 2}$ .)