A magnesium-alloy wire of diameter d = 4mm and length L rotates inside a flexible tube in order to open or close a switch from a remote location (see figure). A torque Tis applied manually (either clockwise or counterclockwise) at end 5, thus twisting the wire inside the tube. At the other end A, the rotation of the wire operates a handle that opens or closes the switch. A torque T 0 = 0.2 N · m is required to operate the switch. The torsional stiffness of the tube, combined with friction between the tube and the wire, induces a distributed torque of constant intensity t = 0.04N − m/m (torque per unit distance) acting along the entire length of the wire. (a) If the allowable shear stress in the wire is T a l l o w = 30 MPa, what is the longest permissible length L m a x of the wire?
A magnesium-alloy wire of diameter d = 4mm and length L rotates inside a flexible tube in order to open or close a switch from a remote location (see figure). A torque Tis applied manually (either clockwise or counterclockwise) at end 5, thus twisting the wire inside the tube. At the other end A, the rotation of the wire operates a handle that opens or closes the switch. A torque T 0 = 0.2 N · m is required to operate the switch. The torsional stiffness of the tube, combined with friction between the tube and the wire, induces a distributed torque of constant intensity t = 0.04N − m/m (torque per unit distance) acting along the entire length of the wire. (a) If the allowable shear stress in the wire is T a l l o w = 30 MPa, what is the longest permissible length L m a x of the wire?
Solution Summary: The author explains the expression for the applied torque on the wire, which is t_0+tL.
A magnesium-alloy wire of diameter d = 4mm and length L rotates inside a flexible tube in order to open or close a switch from a remote location (see figure). A torque Tis applied manually (either clockwise or counterclockwise) at end 5, thus twisting the wire inside the tube. At the other end A, the rotation of the wire operates a handle that opens or closes the switch.
A torque T0 = 0.2 N · m is required to operate the switch. The torsional stiffness of the tube, combined with friction between the tube and the wire, induces a distributed torque of constant intensity t = 0.04N − m/m (torque per unit distance) acting along the entire length of the wire.
(a)
If the allowable shear stress in the wire is T allow = 30 MPa, what is the longest permissible length Lmaxof the wire?
I had a theoretical question about attitude determination. In the attached images, I gave two axis and angles. The coefficient of the axes are the same and the angles are the same. The only difference is the vector basis. Lets say there is a rotation going from n hat to b hat. Then, you introduce a intermediate rotation s hat. So, I want to know if the DCM produced from both axis and angles will be the same or not. Does the vector basis affect the numerical value of the DCM? The DCM formula only cares about the coefficient of the axis and the angle. So, they should be the same right?
3-15. A small fixed tube is shaped in the form of a vertical helix of radius a
and helix angle y, that is, the tube always makes an angle y with the horizontal.
A particle of mass m slides down the tube under the action of gravity. If there is
a coefficient of friction μ between the tube and the particle, what is the steady-state
speed of the particle? Let y
γ
30° and assume that µ < 1/√3.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.