Frictional forces acting on log Application of Newtons Laws

In summary: N For a,In summary, when a rope with a horizontal force of magnitude 706 N is pulled, a log with a mass of 97.8 kg moves at constant speed. The magnitude of the resistive (frictional) force acting on the log is -706 N.For b,In summary, to pull the log with an acceleration of magnitude 1.46 m/s2, a horizontal force of magnitude 706 N must be exerted on the rope. Assuming the resistive force has not changed, the magnitude of the force needed to pull the log at this acceleration is also 706 N.
  • #1
destro47
16
0
A rope is tied to a 97.8 kg log. When you pull on the rope with a horizontal force of magnitude 706 N, the log moves at constant speed.

a) Find the magnitude of the resistive (frictional) force acting on the log. You may assume the frictional forces acting on the log is constant.
N

b) To pull the log with an acceleration of magnitude 1.46 m/s2, what is the magnitude of the horizontal force must you now exert on the rope? Assume the resistive force has not changed.

This one has really stumped me, I don't know where to begin as my physics class has not even covered friction yet. My professor assigned this as part of an online homework assignment for a lecture he did not even give (due to the recent holidays). He doesn't want to lose time because we have a lot of material to cover this semester. This is due tomorrow and it has stumped me all week, someone please HELP!
 
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  • #2
For a,

The sum of the forces in the x direction = ma

Pulling force + Frictional force = m(0) (remember + or - signs depending on direction, friction acts against movement)

I think that should help.


For b, use the same value of the frictional force from part a, then

Pulling force + Frictional force = ma

solve for the pulling force...
 
  • #3
The biggest problem that I had with this problem is that the log moves at constant v, so a=0. How can a force move on object with 0 acceleration? Is this log considered to be in equillibrium even if its moving?
 
  • #4
so basically the answer is -706 N if i understand you correctly:

Pulling Force + Friction Force = m (0)


706 N + f = m (0)

f = 0 -706 N
 

Related to Frictional forces acting on log Application of Newtons Laws

1. What is frictional force and how does it affect objects?

Frictional force is the resistance between two surfaces when they come in contact with each other. It is caused by microscopic irregularities on the surface of the objects. Frictional force can either slow down or prevent movement of an object.

2. How is frictional force related to Newton's Laws of Motion?

Frictional force is directly related to Newton's First Law of Motion, also known as the Law of Inertia. This law states that an object will remain at rest or in motion unless acted upon by an external force. Frictional force acts as an external force and can change the motion of an object.

3. What is the formula for calculating frictional force?

The formula for calculating frictional force is F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force. The coefficient of friction depends on the type of surfaces in contact and the normal force is the force perpendicular to the surface of contact.

4. How does the surface area of an object affect frictional force?

The surface area of an object does not directly affect the magnitude of frictional force, but it can indirectly affect it. A larger surface area means that there is a larger area for frictional force to act on, resulting in a larger overall force. This can be seen when comparing the frictional force of a small ball rolling on a surface versus a large box sliding on the same surface.

5. How can frictional force be reduced or increased?

Frictional force can be reduced by using lubricants or by using smoother surfaces. It can be increased by increasing the normal force or by increasing the coefficient of friction, such as by using rougher surfaces or increasing the weight of an object.

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