Answer:
The torque at the end of the beam is 2.5 Nm
Explanation:
Given;
length of beam, r = 0.5 m
applied force, F = 5 N
The torque at the end of the beam is given by;
τ = F x r
where;
τ is the torque
F is applied force
r is length of the beam
τ = 5 x 0.5
τ = 2.5 Nm
Therefore, the torque at the end of the beam is 2.5 Nm
