Finding Horizontal Force Needed to Move an Object

AI Thread Summary
To find the horizontal force required to move an object, the formula is Fr = μN, where Fr is the resistive force of friction, μ is the coefficient of static friction, and N is the normal force. The normal force can be calculated as N = mg, with m being the mass of the object and g the acceleration due to gravity (approximately 9.81 m/s²). The maximum static friction force is determined by multiplying the coefficient of static friction by the normal force (μs * Fn). For a 6 kg object with a static friction coefficient of 0.42, the normal force is 58.86 N, resulting in a required force of 24.72 N to initiate movement. Understanding these calculations is essential for determining the force needed to overcome static friction.
Sebkl23
Messages
3
Reaction score
0
How can we find the horizontal force required to move an object with the coefficient of static friction and the mass??

I just need a formula please, thankyou
 
Physics news on Phys.org
Sebkl23 said:
How can we find the horizontal force required to move an object with the coefficient of static friction and the mass??

I just need a formula please, thankyou

Fr = μN

Fr - resistive force of friction
μ - coefficient of friction
N - normal force (gravity/applied pressure)
 
a) The friction force only acts in the opposite direction to a horizontal force on an object. The only forces acting on the block described are gravity and the normal force from the table, both of which are in the vertical direction. Therefore, there is no friction force acting on the block.

b) If a horizontal force F acted on the block, the force due to static friction would then begin to act upon the block. This would happen until the force F became greater then the opposing friction force, which would result in the blocks motion. Problem b would be solved by finding the maximum force due to static friction. This is equal to the coefficient of static friction multiplyed by the normal force: μs*Fn.

c) The static friction force equals the force F acting upon the block until F rises greater then the value of the maximum static friction force. So, if half the force needed to overcome the static friction force were applied, 1\2 SF (SF equaling static friction force max), then the static friction force would be an equal and opposite force 1\2 SF as well. This would be written as 1/2*m*g*μs, were m is the mass of the block, g is the acceleration due to gravity: 9.81m\s2, and μs is the coefficient of static friction. (the normal force is equal to the mass of the block multiplied by the acceleration due to gravity)

d) The sum of the horizontal (vertical and horizontal forces are calculated separately) forces equals the mass times the horizontal acceleration of an object as stated in Newtons second law. The two forces acting upon the block are a force F and the kinetic friction force (the kinetic friction force is the friction force acting upon moving bodies, where as the static friction force is upon stationary bodies.) The equation is written F-Kf=m*a, where Kf is the force due to kinetic friction.(Note: the reason the kinetic friction force is subtracted from the horizontal force acting on the block instead of vise versa, is because the movment of the block is in the direction of the force F.) Solving for the acceleration of the block, the mass m is moved to the other side of the equation like this (F-Kf)\m=a or (F-m*a*μs)\m=a. (the kinetic friction force is equal to the coefficient of kinetic friction multiplied by the normal force)

Hope this is what your looking for.
 
Ok so ... let's say I have a 6kg object, I know the coefficient of static force is 0.42, and I want to know what force is required to move the object ...

so I start by finding the normal force of the object by multiplying the mass and force of gravity
which would be: Fn= 6kg X 9.81m/s2= 58.86
Fn= 58.86N

and now I can find the needed force? μf= F/Fn
0.42= F/58.86
F= 0.42×58.86
F=24.72N
so a 24.72N force is needed to move the object?
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top