Presentation on theme: "Physics Montwood High School R. Casao"— Presentation transcript:
1 Physics Montwood High School R. Casao TorquePhysicsMontwood High SchoolR. Casao
2 TorqueAs we have seen over and over again, a force is a push or a pull, and an unbalanced force causes an object to move (to accelerate).All the motion we have dealt with so far has moved objects from one point in space to another. This type of motion involving a displacement is called translational motion.We will now consider another type of motion, rotational motion, where an object moves not from one place to another but rather around some axis of rotation.
4 Torque Forces are also necessary for rotational motion. The rotational effect of a force depends upon its point of application.Torque: is the rotational analog of a force and measures the tendency of a force to rotate a body around a pivot (fulcrum). Actual rotation does not need to occur.
5 Torque A torque is not a force. Pivot point: the point from which the lengths of all torque arms, also called moment arms, are measured.Once a pivot point is chosen, a perpendicular line is drawn to each of the lines along which the force vectors act on the object. Torque arm or moment arm is the perpendicular distance from the pivot point to the line of action of the force.
6 TorqueThe line of action of the force passes through the point at which the force is applied and is parallel to the force.Calculating a torque always requires that you choose one point as your only pivot and calculate all torques relative to it.Torque, the rotational effect of a force is the product of a force and the length of its torque arm.
7 Torque Torque = Force applied · torque arm Mathematically: Unit: m·N or N·mWhen forces are not applied parallel (00) to or perpendicularly (900) to the torque arm, the mathematical expression for the torque is:
8 Torquewhere is the angle between the line of action of the force and a line drawn from the point of application of the force to the pivot.The torque arm is once again the perpendicular distance from the pivot to the line of action of the force.If the distance from the point of application of the force is r, then the torque arm is one side of a right triangle in which r is the hypotenuse.
9 Torque (simplest case) “Lever arm” or torque arm is perpendicular to force.Units: N·m
10 TorqueThe torque for a force applied at the pivot point is always zero. The moment arm for any force applied at the pivot point is zero.Direction of torque:To identify a torque as clockwise or counterclockwise, imagine that the object is free to rotate around a stationary pivot point.Imagine that the force producing the torque is the only force acting on the object.The direction in which the bar would rotate is the direction of the torque.
12 TorqueHow do we quantify torque when force is not perpendicular to the lever arm?Define:Line of Action:The line along the forcevector.Lever Arm:The distance between the line of action and the axis of rotation measured along a line that is perpendicularto both.
13 TorqueZero torque if force isdirected toward center ofrotation.
14 Conditions for equilibrium Mechanics of Rigid BodyEQUILIBRIUM CONCEPTConditions for equilibriumequilibrium = no acceleration, i.e. a particle remains at rest, or if in motion, moves with constant velocity.EQUATIONS OF MOTION OF A RIGID BODYIn the case of a rotation about a fixed axis and O is a point of this axis. In general, if O is a point of a inertial reference systemCONDITIONS FOR EQUILIBRIUM1. The net external force acting on the body must remain zero.2. The net external torque about any point must remain zero. Any point can be considered because all of them are in a inertial reference system (non accelerated).
15 Rotational Equilibrium An object at equilibrium has no net influences to cause it to move, either in translation (linear motion) or rotation. The basic conditions for equilibrium are:
16 Rotational Equilibrium A torque that tends to rotate something counterclockwise is usually defined as positive; a clockwise torque is defined as negative.Rotational equilibrium occurs when the angular acceleration and the sum of the torques is equal to zero.Dynamic rotational equilibrium: angular velocity is constant; angular acceleration is zero.
17 Rotational Equilibrium Static rotational equilibrium: angular velocity and angular acceleration is zero.Three factors determine how effective a torque is in altering the rotational state of an object:the magnitude of the force,the direction of the force, andthe point of application of the force.
18 Problem-Solving TipsAll torques must be calculated about the same pivot.When a problem contains two unknowns, mathematically remove one of the unknowns by selecting it as the pivot. This causes the torque due to the applied force at that location to be zero.
19 Coupled ForcesCouple: two forces of equal magnitude acting in opposite directions in the same plane, but not applied at the same point.Although the forces are in opposite directions, each tends to rotate the body in the same direction about the pivot.The torque of a couple is the product of one of the forces and the perpendicular distance between them.A couple can only be balanced by a second couple; the torques of the two couples must have equal magnitudes but opposite directions.
20 The Lunch Tray ExampleYour fingers and thumb both exert a force on the tray that acts to keep it in equilibrium.Tray: m = 0.2 kg; food: m = 1.0 kg;cup: m = kgWhat is F and T?
22 Determine the components of the reactions at A and B. Mechanics of Rigid Body: StaticsDetermine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A.Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components.A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G.Determine the components of the reactions at A and B.Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.
23 Free-body diagramFirst step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram.Select the extent of the free-body and detach it from the ground and all other bodies.Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body.Include the dimensions necessary to compute the torque arms of the forces.
24 A man raises a 10 kg joist, of length 4 m, by pulling on a rope. Mechanics of Rigid Body.SOLUTION:Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A.A man raises a 10 kg joist, of length 4 m, by pulling on a rope.Find the tension in the rope and the reaction at A.The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R.
25 Equilibrium Example: Ladder WF=875 NWL=355 NdL=8.0 mdF=6.3 mFind the forces that the wall and the ground exert on the ladder.