In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Another definition of torque is the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically
τ
{\displaystyle {\boldsymbol {\tau }}}
or τ, the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted by M.
In three dimensions, the torque is a pseudovector; for point particles, it is given by the cross product of the position vector (distance vector) and the force vector. The magnitude of torque of a rigid body depends on three quantities: the force applied, the lever arm vector connecting the point about which the torque is being measured to the point of force application, and the angle between the force and lever arm vectors. In symbols:
{\displaystyle \tau =\|\mathbf {r} \|\,\|\mathbf {F} \|\sin \theta \,\!}
where
τ
{\displaystyle {\boldsymbol {\tau }}}
is the torque vector and
τ
{\displaystyle \tau }
is the magnitude of the torque,
r
{\displaystyle \mathbf {r} }
is the position vector (a vector from the point about which the torque is being measured to the point where the force is applied),
F
{\displaystyle \mathbf {F} }
is the force vector,
×
{\displaystyle \times }
denotes the cross product, which produces a vector that is perpendicular to both r and F following the right-hand rule,
θ
{\displaystyle \theta }
is the angle between the force vector and the lever arm vector.The SI unit for torque is the newton-metre (N⋅m). For more on the units of torque, see § Units.
In deriving the ##k_{net}## of the given system, it is taken that the extension produced by both springs is equal but their force is different. Therefore ##(k_1+k_2)x=k_{net}x \implies k_1+k_2=k_{net}##.
In absence of pivot, an object rotates around an axis through COM and perpendicular to...
I am trying to size the two motors needed for a robot arm I'm building.
The first motor, which I call the wrist motor, drives a pulley on a fixed axis that connects to a second pulley (of the same size) which is attached to a hand and load.
I can calculate the moment of inertia for the hand and...
TL;DR Summary: I'm stuck trying to find the equation for time period T of a physical pendulum without any calculus using torque.
Hello all.
I am currently writing my IB Physics HL IA (high school physics lab report).
I am investigating the effect of length on the time period of a uniform rod...
Hey Guys,
I'm looking for any help in determining what the estimated torque would be on our mixer. If you know of someone that can help with this please let me know and maybe forward this to them. We are looking for a device (strain gauge) that can display this and we will need to know...
So I started by just figuring out what forces are going to have torque. I know the one heading straight down from the pivot won’t have any and the one going at an angle from the pivot won’t be included in the net torque since it’s at the pivot. The rest of the forces have torque and they are...
I am designing a mini sumo robot which should contain two DC motors. I made a design for some motors which output a stall torque of 4.6 kg-cm each. My wheel radius is 3.4 cm and a depth of 2.2 cm and I plan to make it out of silicone. Wanting as little slip as possible while stalling what could...
We have two gears A and B (left and right). Gear A is driven with a clockwise torque. Why is gear B's torque also clockwise? I would say that if gear B is driven to turn counterclockwise, the torque should be in the counterclockwise direction.
So I have four motors, each capable of producing 1 newton meters of torque. They are all running into a singular drive axle at 90degree offsets. There's no gearbox exchange or power transfer medium. It's axle to axle engagement with a pinon gear. Disregarding any sort of losses that may be...
For this,
I don't understand why they don't have a negative sign as the torque to the friction should be negative. To my understanding, I think the equation 5.27 should be ##I\frac{d \omega}{dt} = -F_{friction}R## from the right hand rule assuming out of the page is positive.
Noting that ##f_k...
TL;DR Summary: When a cube is supported at the fulcrum and remains stationary due to a balloon exerting a force in the opposite direction of its weight.
So the exercise is as follows: We have a homogeneous cube with an edge length of 2 meters, weighing 98N. On the other hand, we have a balloon...
Trying to figure out how much torque an electric motor would need to move a load of 100lbs on a flat surface. The device would use two electric motors attached to a 8 inch diameter wheel (each motor is attached to a 8 inch wheel).
Say I have a bolt and I thread it into a threaded hole. If I torque it down to a certain torque, then use a paint marker and mark the head of the bolt and the surface that the threaded hole is in with a straight line. Then I unfasten the bolt and refasten it into the same exact hole, and torque...
I've already got the answer and the way to solve it (parallelogram), but I'm just wondering why I cannot use the technique I've learned in the lesson torque.
Let's focus on the line AB, if I use what I've learned in torque, the components would be like this:
To find the force component in...
I'm confused because torque increases with increasing moment of inertia (MOI) but angular velocity decreases with increased MOI because of conservation of angular momentum so angular acceleration would also decrease. And then power is the product of torque and angular velocity so with a smaller...
Why did he give torque number 4 zero?
It's not touching the axis of rotation and the angle 90 degrees between them.
I get this:
##\tau## = 1 - 0.8 - 0.4 + 0.4 = 0.2 (C.C.W)
I need help calculating the motor torque requirements for a project I'm working on...the senario is as follows...
Two 5/8" diameter aluminum rods that are 3 feet long and 6" apart from each other. These rods need to rotate a maximum load of 35 pounds at a speed of 25 rotations per minute...
I've inserted a photo of the figure, hope everyone can see it.
SOLUTION:
1. I first solved for the angle of 100 N and 50 N since I need the force that is perpendicular to point A.
>> Angle of 100 N
theta = arctan(3/4)
theta = 36.870 degree
>> Angle of 50 N...
I need some help here. I'm trying to determine how to use the numbers I'm getting for my pony break to determine horsepower at the axle of my engine. Right now I'm getting a reading of six pounds of torque, 4 in off of the center line of the axle at a rate of 30 rpms. Can someone help me...
I have been trying to understand this proof from the book 'Introduction to classical mechanics' by David Morin. This proof comes up in the first chapter of statics and is a proof for the definition of torque.
I don't understand why the assumption taken in the beginning of the proof is...
The total force acting on the pulley is zero so:
F=mg+T1+T2 (1)
Analyzing the torque and angular acceleration about the actual axis of rotation, the axle of the pulley, gives:
τnet=T1R−T2R=Iα (2)
If we analyze about point P, the right edge of the pulley where T1 is applied, we get...
I started by summing the forces and torques to get:
- ma = mg-T
- I*alpha=Tr
I then used a=alpha*r and I=Mr^2 to combine the equations and solved for angular acceleration equals 81.75rad/s^2. Plugging this back into a torque equation I got that the net torque is 1.04Nm. However, the problem...
This is how I interpreted the problem,
a) The net torque about point A is zero. This is because the forces F1 and F2 are equal and opposite, and they act at the same distance from point A. Therefore, they produce torques that cancel each other out..
The force F3 doesn’t does not produce any...
Hallo
Could someone suggest me how to design a barrel cam (for example like in the butterfly valve actuator) to improve the initial torque generated by the movement ?
From what I understand about torque, it is basically the power of the force to cause a change in an object's rotational motion. It is easier to cause this change when the force is applied further from the point of rotation than closer, which is why it is difficult to open a door by pressing a...
Hello all, the short story is, I am basically trying to get the same ( or a little more) “power” out of a larger lawn mower type gas engine ( vertical shaft) as one of the largest (120 ish cc) chainsaws ( ref the Stihl MS 880 or newer 881 at approx 10-12k rpm and 9 Hp)
The parameters are : a...
Hi,
I am running a computational fluid dynamics (CFD) simulation. Supposed I have a symmetrical rigid body in space experiencing torque in the global x,y,z axes. It is stationary at t = 0. I also constrain it to only allow rotations in 3DOFs, and no translation.
It will rotate and I need to...
I need to write an equation for Newton's second law for the above system, where k1=k2 (both springs are the same). The red line represents a bar with m=2kg, l=2m.
I know that I*α = M1 + M2 + M3
If we displace the bar by x, we get the angle of displacement theta.
M1=M2=-k*x
I know that...
Here is a picture of the problem
It is not clear to me how to really prove that the equation for ##\theta(t)## is simple harmonic motion, and what the period of this motion is.
A solution was provided:
We take torques about point B. Note that τ = MgL/2 = Iα so α = (3g)/2L. Everything from here is straightforward.
I don't understand why in this step, you can take torque about B without accounting for a fictitious force due to the acceleration of the Rod.
Thanks for...
The weight of the rack is supported on an axial bearing as seen in the attached pdf below. I have made an attempt to calculate the torque by taking a look at the chain traction force and the required shaft power to make the plates rotate. For the moment of inertia case i don't know how to treat...
I attempted to solve this problem by considering the torque caused by the perpendicular components of the tension and weight with respect to the derrick. $$ Tcos\theta x = Wsin\theta L$$ $$T = \frac L x Wtan\theta$$ Using the principle of virtual work I also arrived at the same answer by...
One image is attached is the question and the other is my attempt. I believe the pillars would have a downward force of 11760N applied to each pillar since they are spaced an equal distance apart. Now the tricky bit is when Torque comes into play. I believe I need to find the distance the car...
^ This is my personal drawing of the diagram, I couldn't take a picture of the actual one. The setup is a pulley wrapped with a cord and mass hangers attached to each end.
My first thought when approaching this problem was to first determine the rotational inertia of the pulley, then use some...
The planet is faster when it is closer to the planet because when it is closer to the planet it has less rotational inertia, and rotational momentum is conserved in this system, so less rotational inertia means a greater angular velocity. This explains why it is slower when it is farther away...
What I did was plug in the outer radius time the force into the torque and then the mass moment of inertia is equal to m*ro^2 so then I plugged in the mass times the radius of gyration squared into I and solved for a but this is not right.
Attempted creating equations for zeros of torque and components of forces in x and y as seen in picture. Got lost with having only variables and the d & 2L for the length of the beam. Not sure how to do the question with two points of contact between the beam and the sign. Is the center center...
Part -B
$$\sum \tau_{cw} = \sum\tau_{ccw}$$
$$\tau_B=\ torque\ of\ the\ beam $$
$$\tau_S =\ torque\ of\ the\ sign\ board$$
$$\tau_C = \ torque\ of\ the \ cable$$
$$\tau_B+\tau_S = \tau_C$$
$$F_B\cdot d_1 + F_S\cdot d_2 = F_C \cdot d_3$$
Since the tension in the left and right chains are evenly...
The net torque about an axis through point A is given by,
If I take the axis of rotation perpendicular to the paper and the solution I arrive would be the following below
Net torque = 30 cos45 x 1.5 - 10 cos30X 3
= 5.829Nm ( counterclockwise)
But the book gives an answer...
Angular Momentum and Torque are defined about a point. But Moment of Inertia of a body is defined about an axis. There are equations which connect Angular momentum and Torque with Moment of Inertia. How will this be consistent? When I say that the torque of a force acting on a body about a point...
Torque limiting extensions act as a torsion device by flexing once a certain amount of torque is applied onto the fastener that they are tightening down. This helps prevent from over tightening the fastener when using an impact wrench. Seems like a great tool that can speed up the process of...
F_{1}d_{1} + F_{2}d_{2} = F_{3}d_{3} + F_{4}d_{4}
m_{1} gR cos 60 + m_{2}gR cos 60 = m_{3}gR cos 60 + m_{4}gR sin 90
m1 = m2= m3= m4= m
R1=R2=R3=R4=R
\sigma\tau = sin 90 - cos 60 = 0.5 Nm.
Have I done this right?
I am wondering if it is possible to calculate either the Kinetic Energy or Rotational Kinetic Energy of an object if we have the Power (kW), Torque (Nm), and Speed (RPM) of the object.
Hello everyone,
I am reading a book about wind power turbines and found a calculation. I tried it myself and the numbers doesn't match.
Here is an image of the data and result.
I don't get to the same torque.
And also what do you think about the moment of inertia value?
Before the data...
Hello, I am trying to figure the strength (in lbs) of a strap needed to attach 2 Rolls together without breaking. Each wheel has a weight of 1000lbs and a diameter of 30in. If there is required information missing, let me know.