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How do you find the normal force if the given are mass and tension?
Wiki User
∙ 11y ago
First . . . look at the drawing, and keep looking until you understand how
the normal force is related to the tension.
Here's a hint that may or may not help: Remember that if nothing is moving,
then the sum of all the forces at any point is zero. And whatever you do,
don't forget about gravity, and its effect on the mass..
Wiki User
∙ 11y ago
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How do you find the normal force of a 25 kg object being pulled by a rope at a tension of 120 N along a frictionless surface at an angle of 30 degrees?
I could be totally wrong but I believe you must take into account the forces on the object in the vertical direction. In this case, the object is being pulled by a rope with a tension of 120 N. The vertical force would therefore be 120 sin(30) Normally W = [F normal] with no vertical force. However, since there is a vertical force: W = [F normal] + [120 sin (30)] (25 kg X 9.8 m/s2) = [F normal] + [120 sin (30)] 245 N - 60 N = [F normal] 185 N = [F normal] I apologize if this is incorrect as I haven't been in a physics class in over 4 years. Good luck!
How can you find force when given distance and speed?
You cannot. Force is mass times acceleration. You have neither.
How do you find force given weight and work?
You cannot. You do not have the necessary information.
How can you find force when given time velocity and mass?
Force equals mass times acceleration.
How can you find force when mass and velocity are given?
You cannot. Force = Mass*Acceleration or Mass*Rate of change of Velocity.
How do you find pressure if force and area is given?
force= pressure*area
Why is there a relationship between velocity-squared and tension but not velocity and tension?
There is, the equation given is veloctiy = sqr root of (Tension/mew) where mew is a constant for the length of string and is given by mew = mass/length. by rearranging to find mew, we get either velocity2/Tension giving 1/mew or we can get Velocity/sqr root of Tension giving 1/sqr root of mew.
How do you find the normal force of a 25 kg object being pulled by a rope at a tension of 120 N along a frictionless surface at an angle of 30 degrees?
I could be totally wrong but I believe you must take into account the forces on the object in the vertical direction. In this case, the object is being pulled by a rope with a tension of 120 N. The vertical force would therefore be 120 sin(30) Normally W = [F normal] with no vertical force. However, since there is a vertical force: W = [F normal] + [120 sin (30)] (25 kg X 9.8 m/s2) = [F normal] + [120 sin (30)] 245 N - 60 N = [F normal] 185 N = [F normal] I apologize if this is incorrect as I haven't been in a physics class in over 4 years. Good luck!
How do you find pressure given force and area?
P = F/A, so Area = Force/Pressure
How can you find force when given distance and speed?
You cannot. Force is mass times acceleration. You have neither.
How do you find force given weight and work?
You cannot. You do not have the necessary information.
How can you find force when given time velocity and mass?
Force equals mass times acceleration.
How can you find force when mass and velocity are given?
You cannot. Force = Mass*Acceleration or Mass*Rate of change of Velocity.
How do you find work done if force and mass are given?
You cannot. You do not have the necessary information.
Where would you expect to find tension?
Normal fault
How do you find force in science?
It depends on the data you are given, but one would be F = ma.
How do i find a Normal reaction force on an angle?
I'm not sure that I understand the question but if you are asking how to find the normal component to a force that is acting on an angle then you should break up the force vector into two components that act at right angles to each other and where one is 'normal' to the (surface of) the object. Normal in this case means "at right angles to a tangent" (I assume that the most common case in dynamics is for the extension of that 'normal' vector to go through the center of gravity of the object).
