Applications of Newtons Laws (friction)

In summary, the problem involves finding the normal force and tension in a strap at a 45 degree angle above the horizontal, as a 18-kg suitcase is pulled with constant speed on a surface with a coefficient of kinetic friction of 0.38. The equations used are N + mg + fk + f = ma and Y comp: N-mg + Fsin(theta) = 0, with two unknowns N and F.
  • #1
sucksatphysic
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Homework Statement



At the airport, you pull a 18-kg suitcase across the floor with a strap that is at an angle of 45 deg above the horizontal. Find

a) Normal force and

b) the tension in the strap, given that the suitcase moves with constant speed and that the coefficient of kinetic friction b/w the suitcase and the floor is 0.38.

Homework Equations



N + mg +fk+ f = ma

I'm not sure if this equation is correct because they don't give you an initial force...

Y comp: N-mg + Fsin(theta) = 0


The Attempt at a Solution



I just need to know if my equations are right because the don't give an initial force. help?
 
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  • #2
I like "Y comp: N-mg + Fsin(theta) = 0".
I think you'll have to write a force equation like that for the horizontal part, too.
Likely you'll have two unknowns N and F, which you can solve for when you have the two equations.
 
  • #3


I can confirm that your equations are correct. The equation N + mg + fk + f = ma is a valid representation of Newton's Second Law, where N represents the normal force, mg represents the weight of the suitcase, and fk and f represent the kinetic friction force and the applied force, respectively. Since the suitcase is moving at a constant speed, the acceleration is zero and the equation can be simplified to N + mg + fk = 0.

For part a, to find the normal force, we can use the Y component equation N - mg cos(theta) = 0, where theta is the angle of 45 degrees. Solving for N, we get N = mg cos(theta). Plugging in the values of m (18 kg), g (9.8 m/s^2), and cos(theta) (cos(45) = 0.707), we get N = 124.92 N.

For part b, to find the tension in the strap, we can use the X component equation N - mg sin(theta) - fk = 0. Solving for fk, we get fk = N - mg sin(theta). Again, plugging in the values, we get fk = 124.92 N - (18 kg)(9.8 m/s^2)(sin(45) = 0.707) = 49.97 N. This is the tension in the strap required to maintain a constant speed.

In conclusion, Newton's Laws, specifically the equations for forces and components, can be applied to solve for the normal force and tension in the strap in this scenario. The coefficient of kinetic friction also plays a crucial role in determining the required tension in the strap. This application of Newton's Laws in understanding and predicting the motion of objects is a fundamental aspect of physics and has many practical applications in various fields.
 

FAQ: Applications of Newtons Laws (friction)

1. What is friction and how does it relate to Newton's Laws?

Friction is the force that opposes motion between two surfaces in contact. It is directly related to Newton's Laws of Motion, specifically the First and Second Laws. The First Law states that an object at rest will remain at rest unless acted upon by an external force, and an object in motion will remain in motion unless acted upon by an external force. Friction is the force that acts against the motion of an object, causing it to slow down or come to a stop. The Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Friction is one of the forces that can act against an object's acceleration, thus affecting its motion.

2. What are some real-life examples of friction in action?

Some common examples of friction in everyday life include walking on a rough surface, rubbing your hands together to create heat, and slowing down a car by applying the brakes. In all of these scenarios, friction is acting on the objects involved, either to slow down motion or to create heat.

3. How can friction be both helpful and harmful?

Friction can be helpful in allowing us to walk without slipping, in providing traction for vehicles, and in allowing us to grip objects. However, it can also be harmful in causing wear and tear on surfaces, creating heat that can damage machinery, and causing discomfort or blisters when walking or using tools for extended periods of time.

4. How does the coefficient of friction affect the amount of force needed to move an object?

The coefficient of friction is a measure of the roughness of the surfaces in contact. The rougher the surfaces, the higher the coefficient of friction and the more force is needed to move an object. Conversely, smoother surfaces have a lower coefficient of friction and require less force to move an object.

5. How can we reduce friction in order to increase efficiency?

There are several ways to reduce friction in order to increase efficiency. One way is to use lubricants, such as oil or grease, to create a barrier between surfaces and reduce the amount of contact and friction. Another way is to use smoother materials for surfaces that come in contact, such as using ball bearings in machinery. Additionally, reducing the weight of objects can also reduce the force needed to overcome friction and increase efficiency.

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