A ring with radius \(4~ m\) hinged at point P. Two forces \( F_{1} \) and \( F_{2} \) each of magnitude \(10~ N\) act tangentially at the surface of the ring. Find the ratio of the torques produced by two forces \(F_1\) to \(F_2\) is
A ring with radius \(4~ m\) hinged at point P. Two forces \( F_{1} \) and \( F_{2} \) each of magnitude \(10~ N\) act tangentially at the surface of the ring. Find the ratio of the torques produced by two forces \(F_1\) to \(F_2\) is
Given,
Radius of ring \(r\) = \(4~m\)
Force \(F_1\) = \(10~N\)
Force \(F_2\) = \(10~N\)
Torque due to force= Force \(\times\) perpendicular distance between line of action of force and axis of rotation
Let perpendicular distance of point of rotation P from the Force \( F_{1} \) is \( d_{1} =8~m \) while that of force \( F_{2} \) is \( d_{2} = 4~m \)
Torque \(T_1\) produced by \(F_1\) = \(d_1\times F_1\) = \(8\times10\) = \(80~Nm\)
Torque \(T_2\) produced by \(F_2\) = \(d_2\times F_2\) = \(4\times10\) = \(40~Nm\)
Ratio of torques \(\frac{T_1}{T_2}~=~\frac{2}{1}\)
Given,
Radius of ring \(r\) = \(4~m\)
Force \(F_1\) = \(10~N\)
Force \(F_2\) = \(10~N\)
Torque due to force= Force \(\times\) perpendicular distance between line of action of force and axis of rotation
Let perpendicular distance of point of rotation P from the Force \( F_{1} \) is \( d_{1} =8~m \) while that of force \( F_{2} \) is \( d_{2} = 4~m \)
Torque \(T_1\) produced by \(F_1\) = \(d_1\times F_1\) = \(8\times10\) = \(80~Nm\)
Torque \(T_2\) produced by \(F_2\) = \(d_2\times F_2\) = \(4\times10\) = \(40~Nm\)
Ratio of torques \(\frac{T_1}{T_2}~=~\frac{2}{1}\)
