A uniform metal rod, with a mass of 3.6 kg and a lengthof 1.2 m, is attached to

Question

A uniform metal rod, with a mass of 3.6 kg and a lengthof 1.2 m, is attached to a wall by a hinge at its base. Ahorizontal wire bolted to the wall 0.60 m above the base ofthe rod holds the rod at an angle of 30 degrees above thehorizontal. The wire is attached to the top of the rod. Find: A) Tension on the wire B) Horizontal Force C) Vertical Force The current Q/A solution listed isn't helping meunderstand the process. If someone could include a quick diagram itwould help out a lot! A uniform metal rod, with a mass of 3.6 kg and a lengthof 1.2 m, is attached to a wall by a hinge at its base. Ahorizontal wire bolted to the wall 0.60 m above the base ofthe rod holds the rod at an angle of 30 degrees above thehorizontal. The wire is attached to the top of the rod. Find: A) Tension on the wire B) Horizontal Force C) Vertical Force The current Q/A solution listed isn't helping meunderstand the process. If someone could include a quick diagram itwould help out a lot!

Explanation / Answer

When dealing with problems involving rigid objects that have apoint of (possible) rotation, it's useful to look at the torques aswell as the forces. Here are the main things to notice: 1. The rod is not moving, so its acceleration is zero. That meansthat the net force on it is zero (i.e. all the forces cancel eachother). 2. The rod is not rotating, so its _angular_ acceleration is zero.That means the net _torque_ on it is zero (all torques cancel eachother). For part (a), figure the torques first. Use the hinge as therotation point. For convenience (since I don't have a diagram),assume that the wall is on the right, with the rod & wireextending to the left. * There is a counter-clockwise torque due to the weight of the rod.You can assume that this force acts on the rod's center of mass,which (since they said it's "uniform") is right in the center ofthe rod, i.e. 0.6 meters from the end. Draw a diagram, and you'llsee that the lever arm (perpendicular distance from the line offorce to the hinge) is: (0.6m)(cos(30)). So the amount of thistorque is: (1): _1 = (3.6kg)(g)(0.6m)(cos(30)) * There is a clockwise torque due to the wire. The force is "T"(the unknown tension); and the lever arm distance is 0.60 m(because they said the wire is bolted 0.60 m above the hinge). Sothe amount of this torque is: (2): _2 = (T)(0.60m) * There is another force on the rod, namely the wall pushing on therod at the location of the hinge. However since that force passesright through the rotation point, the lever arm distance is zero,so it does not contribute any torque. So, we know that the clockwise torques must exactly cancel out thecounter-clockwise torques, which means _1 = _2. So, setEquation (1) equal to Equation (2), and solve for "T". Now for part "b": Remember that the total forces cancel out, which also means thatthe _horizontal_ forces and the _vertical_ forces also_individually_ cancel out. The horizontal forces on the rod are: * The tension "T" (calculated above), pulling to the right; * The horizontal force "Fx" exerted by the hinge, pushing to theleft. The horizontal forces cancel out so: Fx = T The vertical forces are: * The weight of the rod (3.6kg)(g), acting down; * The vertical force "Fy" exerted by the hinge, acting upward. The vertical forces cancel out, so: Fy = (3.6kg)(g)

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