Physics 6A
Torque is what causes angular acceleration (just like a force causes linear acceleration)
Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Pivot Point F A F C F B
Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Force B will tend to rotate the bolt clockwise, which will tighten it. Pivot Point F A F C F B
Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Force B will tend to rotate the bolt clockwise, which will tighten it. Notice that force A will tend to rotate the bolt counter-clockwise, loosening it. What does force C do? Pivot Point F A F B F C
Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Force B will tend to rotate the bolt clockwise, which will tighten it. Notice that force A will tend to rotate the bolt counter-clockwise, loosening it. What does force C do? Force C doesn t cause any rotation at all there is no torque generated by force C. Why not? Pivot Point F A F B F C
Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Force B will tend to rotate the bolt clockwise, which will tighten it. Notice that force A will tend to rotate the bolt counter-clockwise, loosening it. What does force C do? Force C doesn t cause any rotation at all there is no torque generated by force C. Why not? Force C points directly at the pivot point no torque is created in this case. Pivot Point F A F B F C
Formula for torque: F r sin( ) This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied).
Formula for torque: F r sin( ) Torque This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). Pivot Point r θ F A Take a look at the diagram r and θ are shown for force A.
Formula for torque: F r sin( ) Torque This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). Pivot Point r θ F A Take a look at the diagram r and θ are shown for force A. There are 2 ways to interpret the formula. If you group the Fsin(θ) together, that represents the component of the force that is perpendicular to the radius. To get the most torque, the force should be applied perpendicular (can you see why from the formula?)
Formula for torque: F r sin( ) This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). Pivot Point Lever Arm F A θ r Take a look at the diagram r and θ are shown for force A. There are 2 ways to interpret the formula. If you group the Fsin(θ) together, that represents the component of the force that is perpendicular to the radius. To get the most torque, the force should be applied perpendicular (can you see why from the formula?) The other option is to group the rsin(θ) together and call it the lever arm for the force. Think of this as the shortest distance from the pivot point to where the force is applied. This is the effective radius of the force. Again, to get maximum torque the angle should be 9.
Example: Find the torque of each force shown with respect to the pivot point at the left end of the 2m long rod. F 1 is applied at the right end, and F 2 is at the center. 12 F 2 =3N We can simply use our definition of torque here. 1 2 F r sin( ) (2N) (2m) sin(5 ) (3N) (1m) sin(12 ) 3.6N m 26.N m 5 F 1 =2N Notice the sign convention: Counter-clockwise torque is positive. Clockwise torque is negative.
We mentioned earlier that torques produce angular accelerations. We have a formula for this relationship: I This is really just Newton s 2 nd law applied to rotational motion. The moment of inertia, I, takes the place of the mass, and we use angular acceleration instead of linear.
Example: Find the angular acceleration of the 2m long, uniform rod (mass=5kg) when it is subject to the 2 forces shown. F 1 is applied at the right end, and F 2 is at the center. 12 F 2 =3N This is just like the last problem, so we can use the results here. We need to add up all the torques on the rod, then solve for. 1 I 2 3.6N.69 ( 1 3 m rad 2 s ML 2 )( ) 26.N Look up this formula for the moment of inertia of a rod, with the axis at the end. m ( 1 3 ) (5kg) (2m) 2 5 F 1 =2N
Static Equilibrium Sometimes an object is subject to several forces, but it does not accelerate. This is when the object is in equilibrium. We have done problems like this before, but we neglected the rotational motion. To incorporate this, we simply need to add a torque formula to our typical force formulas. Here s an example: 1.48 from textbook: A uniform beam 4m long and weighing 25N carries a 35N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. 6 1.5m
1.48 from textbook: A uniform beam 4m long and weighing 25N carries a 35N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. We need to draw a diagram of all the forces, then write down force and torque equations: F x F y 6 H y T H x 1.5m 25N 35N T=Tension in wire H x and H y are the components of the force that the hinge exerts on the beam.
1.48 from textbook: A uniform beam 4m long and weighing 25N carries a 35N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. We need to draw a diagram of all the forces, then write down force and torque equations: H H F x x x T T x cos(3 ) 6 H y H x T 3 1.5m This could also be sin(6) We will save this equation and come back to it later. 25N 35N T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam.
1.48 from textbook: A uniform beam 4m long and weighing 25N carries a 35N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. We need to draw a diagram of all the forces, then write down force and torque equations: H H F y y y T T y 25N sin(3 ) 35N 25N 35N 6 H y H x T 3 1.5m This could also be cos(6) We will save this equation and come back to it later. 25N 35N T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam.
1.48 from textbook: A uniform beam 4m long and weighing 25N carries a 35N weight 1.5m from the far end, as shown. It is supported by a hinge at the wall, and a metal wire running from the wall to the far end. Find the tension in the wire, and find the horizontal and vertical components of the force that the hinge exerts on the beam. We need to draw a diagram of all the forces, then write down force and torque equations: Before we can fill in the torque equation we need to choose a pivot point. A convenient choice is where the hinge attaches to the beam. This simplifies the torque equation because the 2 unknown hinge forces will not create any torque about that point. 6 H y H x T 3 1.5m Also, remember the sign convention clockwise torques are negative and counterclockwise positive. Pivot point here 25N 35N T (25N) 6875N (2m) (35N) (2.5m) (T)(4m) sin(3 ) Now we can go back and substitute this value into the other equations to find the hinge forces. T=Tension in wire H x and H y are the components of the force the hinge exerts on the beam. H H x x T cos(3 ) 5954N H H y y T sin(3 ) 2564N 25N 35N