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How can you find the coefficient of friction force only given mass and theta?
Wiki User
∙ 10y ago
The coefficient of friction is the tangent of the angle theta where the angle is measured from horizontal when the mass first starts to slip
Wiki User
∙ 10y ago
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If tansqtheta plus 5tantheta0 find the value of tantheta plus cottheta?
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
What is cos theta multiplied by csc theta?
It is cotangent(theta).
What is cosine squared theta equal to'?
Cosine squared theta = 1 + Sine squared theta
Express csc theta in terms of cot theta theta is in quadrant 3?
It is -sqrt(1 + cot^2 theta)
Letter after theta?
The letter after Theta is Iota.
How can you find the theta if the given is just the coefficient friction and the mass?
If by theta you mean the angle at the base of slope on which is the body laying, and you want to calculate minimal theta for which the blocks starts to slide: Let's first calculate: weight: Q = mg force normal to the slope: N = Q cos theta = mg cos theta force tangent to the slope: F = Q sin theta = mg sin theta force of friction: T = fN = fmg cos theta, where f is coefficient of friction The body will start to move downwards, when T = F, or: fmg cos theta = mg sin theta which after simplyfying becomes: f cos theta = sin theta, f = sin theta / cos theta f = tan theta Therefore, theta = arc tan f As you see, the angle only depends on friction coefficient f. (If that's not a problem you asked to be solved, edit your question please to precisely state what needs to be calculated)
How to find Force of friction around circle?
T1/T2=e^(mu*theta)where T1/2 are the tensions in the circlemu is the coefficient of frictiontheta is the angle of the circle in contact with the rope.
Can you predict the angle at which an inclined plane can be inclined if an object is placed on it and all you know is the coefficient of friction between the object and the surface of the plane?
The frictional force that opposes the motion of an object on an inclined plane is given by the formula:Ffriction = (mu)N,where mu (the Greek letter mu) is the coefficient of friction, and N is the Normal force, which is the force equal and opposite to the component of the object's weight perpendicular to the surface of the incline.The Normal force will be equal to Wcos(theta), where W is the weight of the object (W= mg) and theta is the angle of the incline.When motion down the plane is impending (that is, a split second before the friction is overcome and the object starts to slide down the plane), Ffriction is equal and opposite to the component of the weight parallel to the surface of the plane. That component is equal to Wsin(theta).So, what does that give us?We know that(1) Ffriction = Wsin(theta)(2) Ffriction = (mu)N(3) N = Wcos(theta)Substituting for N in equation (2) gives usFfriction = (mu)Wcos(theta).Equating equ. (1) and (2) gives usmuWcos(theta) = Wsin(theta).Solving for mu gives usmu = sin(theta)/cos(theta)mu = tan(theta)theta = tan-1(mu) or theta = arctan(mu)So, the arctangent of mu is the angle of incline.(I guess I coulda just said that right from the beginning.)
What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.78?
coefficient of friction = 0.8 tan theta use your calculator : 2nd tan ( 0.8) = 38.66 degrees
How do you measure friction when not knowing the normal force?
Well you can always work out the normal force... if the object is on a horizontal surface... The normal force is equal to the mass of the object × gravity (9.81). If the object is on an incline, you have to get the component of weight which is equivalent to the Normal, in most cases it is Normal = mass × gravity × cos(theta), theta being the angle of inclination.
What is the normal force equation?
mgcos(theta)
What is the equation for normal force?
mgcos(theta)
Condition where tangential force is considered almost horizontal?
when sin theta almost equals to theta in radians
What is the value of theta such that a 900 newton block is in limiting equilibrium with mu coefficient of friction equal to 1.3?
Note that I will be using x as a replacement for theta, and m as a replacement for the coefficient of friction.When a block on a ramp is in limiting equilibrium, it means it is literally on the edge of moving forwards. In other words, the force pushing it forwards and backwards cancel each other out. Now we have:the weight of the block= 900NReaction force R (the force perpendicular to the ramp acting against the blocks weight)= 900cos(x)Force down the block = 900sin(x)Force up the block = mR = 1.3*900cos(x)As stated earlier the forces up and down the block need to cancel each other out, in other words Force up-Force down=0 giving1.3*900cos(x) - 900*sin(x) = 0=> 1.3*900cos(x) = 900sin(x)=> 1.3 = tan(x)=> x = arctan(1.3)=> x = 52.4314 degrees or 0.9151 rad
How do you get the csc theta given tan theta in quadrant 1?
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
Force times distance is?
A - WORKWork = F.s cos (theta)
If a 1-N force accelerates a 1-kg mass at 1 meter per second squared in a zero-gravity and zero-friction environment how do you calculate its motion in the presence of gravity and friction?
Your ideal scenario describes the near-perfect vacuum and weightlessness of space. As you said, a one-kilogram mass will accelerate at one meter per second squared if a net force of one newton is applied to it. But you wish to know what happens under real-world conditions, such as the existence of friction (air resistance, rolling friction, and sliding friction) and gravity. If the motion is in the horizontal plane and at low speeds, air resistance is negligible. Friction between surfaces and rolling friction are significant, however, and will act against any force used to accelerate an object. Gravity plays a role in the consideration of friction, as well, inasmuch as the frictional force is proportional to the Normal Force, which is related to the object's weight.1 For motion in two directions, such as the path of a kicked ball or the trajectory of a bullet fired from a gun, air resistance plays a large role as does the acceleration of gravity. Friction and the effects of gravity play a major role in calculations involving the motion in the vertical direction, whereas the acceleration of gravity plays no role in the calculations of the motion in the horizontal direction.2Basically, keep in mind that weight is a force directed downward. When determining the net force acting on an object, its weight is just one of the forces acting on it. 1. For an object resting on a flat surface with zero incline, the Normal Force is equal and opposite to the object's weight (W = mg). The frictional force is calculated to be uN, where u (the Greek letter mu) is the coefficient of friction. The frictional force opposes the direction of motion. For objects on an inclined plane, the Normal force is equal to W*cos(theta), where theta is the angle of incline. 2. Weight is a force perpendicular (orthogonal) to horizontal motion and, hence, can have no effect upon it.
