What is the maximum load for this crane at an angle of 31.0 ◦ with the horizontal? Answer in units of N.

Question

7416.26 N*M as the torque capacity.

What is the maximum load for this crane at an angle of 31.0 ◦ with the horizontal? Answer in units of N.

Answer 1

What is the maximum load for this crane at an angle of 31.0 ◦ with the horizontal? Answer in units of N.

7416.26 N*M as the torque capacity.

Torque is given by the equation

T = F * D = F * L * cos(∅) (1)

where T = the torque (7416.26 Nm as max for this case)

F = the force exerted (max force for a max torque)

D = the distance (m) from the axis of rotation, perpendicular to the force

L = the moment are of rotation (length of the crane boom, in this case)

∅ = the angle between D and L (31^o in this case)

("... for this crane ..." suggests that a missing diagram shows the length of the boom.)

Solve equ (1) for the F_max and plug in the numbers.