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.
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...
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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...
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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.
Greetings
I need to identify the elements of the following safety cluches: we have three different clutches.
I lready identified most of them exept for number 14
Any help would be appreciated !
thank you!
1- output shaft
2-friction surface
3-preloaded spring
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I am trying to determine if a clamp holding a lever with a mass at one end can experience zero or reduced torque either by balancing the forces or converting the rotational torque into another form such as linear, I’ve looked at counterbalance with Steadicams etc. but I believe this just...
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The justification for N1=f and N2=W which I have so far read is that it is...
Question regarding finding the magnitude of Torque of a system of two masses attached to a rod rotating at angle of θ degrees with respect to a vertical
Hello,
Assuming the dome will not spiral away using a single motor and is set up in a manner such that it would rotate in place, I want to find the minimum torque needed to rotate a fairly heavy dome if a motor was located on the dome's inner face.
I understand that this is an issue that...
So, this problem is going to require some explaining..
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Hello everyone!
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I need to tighten the nut for a trailer hitch ball to 250 ft lbs but I don't have a torque wrench that can do it.
I read on a different forum that an alternate old school method to get 450 ft lbs of torque is:
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So in that forum's example, have a 180 lb person stand on...
Hi,
I have an object sitting on the ground, with a coefficient of friction (COF) of 0.3.
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Hi all
I am Belgian and I have difficulty with english formulas.
I'm trying to calculate the power and torque required of a motor to rotate a load.
Here is the problem:
I have a motor (axis up) on which is fixed a cylindrical load of 15Kg. The diameter of this load is 400mm
Calculate the power...
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Hello everyone,
I am trying to design a fixture that can clamp onto and turn an equipment 180 degrees for maintenance.
I am trying to figure out what torque will be required at the hand wheel to turn it.
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In the above...
Hi everyone.
In most cases when torque and lever arm are being discussed, it is the angle between the force and the arm that matters (like in the following picture).
However, non of the articles that I have read so far have mentioned anything about the distance between force and fastener being...
1) Since the rod is uniform, with mass m and length l, it has a linear mass density of ##\lambda=\frac{m}{l}##, so ##I_{rod_O}=\int_{x=r}^{x=r+l}x^2 \lambda dx=\frac{\lambda}{3}[(r+l)^3-r^3]=\frac{\lambda r^3}{3}[(1+\frac{l}{r})^3-1]=\frac{1}{3}mr^2[3+\frac{3l}{r}+\frac{l^2}{r^2}].##...
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Hello,
i have tried to calculate the acceleration (COM) of the cylinder (even though in the question they asked about the angular acceleration) and the answer is:
𝑎(𝑐𝑜𝑚)=𝐹(𝑟/𝑅−𝑐𝑜𝑠(𝑡ℎ𝑒𝑡𝑎))/(𝐼/𝑅2+𝑀)
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