Gareth56 11 Posted January 26, 2014 In the example below it's the length of the rod (4m) that is taken as the lever arm to calculate the torques around the top where the shark and tension rope is attached. Can I ask why? Because in a ladder against the wall [torque] problem the length of the ladder isn't taken as the lever arm it's the perpendicular distance from the pivot point (usually where the leader meets the ground) to a point on the ladder. Thanks. 0 Share this post Link to post Share on other sites

swansont 7443 Posted January 26, 2014 in a ladder against the wall [torque] problem the length of the ladder isn't taken as the lever arm it's the perpendicular distance from the pivot point (usually where the leader meets the ground) to a point on the ladder. I think that's taking a shortcut; T = rf sin(theta). If you take the projection of the ladder that gives you r*sin(theta) 0 Share this post Link to post Share on other sites

Gareth56 11 Posted January 26, 2014 (edited) I think that's taking a shortcut; T = rf sin(theta). If you take the projection of the ladder that gives you r*sin(theta) Thanks for that. I think I understand now how the torque about the shark is taken. You drop a line down from the shark's weight to the base then calculate the lever arm. Torque shark = 10000 x 4 x sin 30. Where I'm having difficulty is seeing where the Tension Torque = T x 4 x sin 80 comes from. I'm lead to understand that it's the perpendicular component of the force that acts on the pivot point is the one that counts. Edited January 26, 2014 by Gareth56 0 Share this post Link to post Share on other sites

swansont 7443 Posted January 26, 2014 Where I'm having difficulty is seeing where the Tension Torque = T x 4 x sin 80 comes from. I'm lead to understand that it's the perpendicular component of the force that acts on the pivot point is the one that counts. What's the angle between the boom and the line providing the tension? 0 Share this post Link to post Share on other sites

J.C.MacSwell 439 Posted January 27, 2014 (edited) In the example below it's the length of the rod (4m) that is taken as the lever arm to calculate the torques around the top where the shark and tension rope is attached. Can I ask why? Because in a ladder against the wall [torque] problem the length of the ladder isn't taken as the lever arm it's the perpendicular distance from the pivot point (usually where the leader meets the ground) to a point on the ladder. Thanks. Assuming the weight of the rod to have no significance: As you can see from the fact that the rod is pinned, not fixed, there is no torque on the rod, it can only be in tension or, in this case, compression. Edited January 27, 2014 by J.C.MacSwell 0 Share this post Link to post Share on other sites

Gareth56 11 Posted January 27, 2014 Never mind, it's obviously a very challenging torque problem. By the way swansont the angle is 80deg!!! Thanks anyway. 0 Share this post Link to post Share on other sites

swansont 7443 Posted January 27, 2014 Never mind, it's obviously a very challenging torque problem. By the way swansont the angle is 80deg!!! Thanks anyway. That's where the sin80 comes from. And the rod is 4m long. 0 Share this post Link to post Share on other sites

J.C.MacSwell 439 Posted January 27, 2014 Never mind, it's obviously a very challenging torque problem. By the way swansont the angle is 80deg!!! Thanks anyway. Do a free body diagram of the forces at the top of the rod 0 Share this post Link to post Share on other sites

studiot 2114 Posted January 27, 2014 In the example below it's the length of the rod (4m) that is taken as the lever arm to calculate the torques around the top where the shark and tension rope is attached. Can I ask why? Because in a ladder against the wall [torque] problem the length of the ladder isn't taken as the lever arm it's the perpendicular distance from the pivot point (usually where the leader meets the ground) to a point on the ladder. I'm not at all sure what you mean by this question. None of the forces involved exert any moment whatsoever about the top of the boom arm where the shark and tension rope is attached, for the very simple reason that the line of action of all three forces acting passes through this point. Three forces? Yes the boom is a body in equilibrium under the action of three forces, the condiftion for which is that they meet at a point. So MacSwell is correct that the hinge reaction force is along the boom since it must pass through the point where the shark suspension and support rope meet ie the top end. Equally the hinge reaction exerts no moment (american=torque) on the hinge since it passes through it. However the weight of the shark and the support rope tension both exert a moment about the hinge, given by simple trigonometry. Drop perpendiculars from the hinge to the shark (horizontal) and the distance is 4 cos (60) and from the hing to the support rope where the distance is 4sin(80) So the clockwise moments (due to shark weight) are 10000 . 4cos60 = anticlockwise moments (due to rope) Tsin(80) This can easily be solved for T 0 Share this post Link to post Share on other sites

J.C.MacSwell 439 Posted January 28, 2014 So the clockwise moments (due to shark weight) are 10000 . 4cos60 = anticlockwise moments (due to rope) Tsin(80) This can easily be solved for T T being the tension on the rope (just in case it is confused with representing torque) and I think the right side of the equation needs the 4m as well, or alternatively cancelled out on the left... 1 Share this post Link to post Share on other sites

studiot 2114 Posted January 28, 2014 Good catch, thanks. 0 Share this post Link to post Share on other sites