Answer:
C.[tex]\frac{25}{7}[/tex]m
Explanation:
We are given that
Weight of board=w=10 N
Length of board=L=5 m
Tension in the string=T=3 N
Applied upward force=F=7 N
We have to find the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.
Let r be the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.
The board is uniform therefore, the center of board is the mid- point of board.
Therefore, the lever arm of weight=[tex]r_1=\frac{L}{2}=\frac{5}{2}m[/tex]
Now, the torque exerted by the weight of the board
[tex]\tau_1=Force\times perpendicular\;distance=10\times \frac{5}{2}=25 N[/tex]
The torque exerted by applied force=[tex]\tau_2=7\times r=7r[/tex]
In static equilibrium
The sum of rotational forces=0
[tex]\tau_1+\tau_2=0[/tex]
The two rotational force act in opposite direction therefore,
[tex]\tau_2=-7r[/tex]
Substitute the values
[tex]25-7r=0[/tex]
[tex]7r=25[/tex]
[tex]r=\frac{25}{7}[/tex]m
Hence, option C is true.
