Help with work, angles with friction

In summary, you do 457.2 J of work to move a 4.2 kg chair 9.49 m across the floor at a constant velocity by pushing down on the back of the chair at some angle. If the coefficient of kinetic friction between the chair and the floor is 0.75, what is the angle (relative to the floor) at which you are pushing?
  • #1
Stingarov
22
0
You do 457.2 J of work to move a 4.2 kg chair 9.49 m across the floor at a constant velocity by pushing down on the back of the chair at some angle. If the coefficient of kinetic friction between the chair and the floor is 0.75, what is the angle (relative to the floor) at which you are pushing?

Says the answer is 25.6 Degrees, but I'm unsure even the right ways to start this problem.

It's not a homework question, just curious if anyone can show me the steps they would use to solve this? Dying to find out.
 
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  • #2
Identify the forces acting on the chair and draw a free body diagram. What does the constant velocity part tell you about the net force acting on the chair?
 
  • #3
F friction (Uk * Fn) and Applied Force, correct?
 
  • #4
Stingarov said:
F friction (Uk * Fn) and Applied Force, correct?

Correct. Now, what is your next step?
 
  • #5
W = F * d,

457.2 = F * 9.49

48.18 = F ? but I'm uncertain where to go from here with it.
 
  • #6
Stingarov said:
W = F * d,

457.2 = F * 9.49

48.18 = F ? but I'm uncertain where to go from here with it.

Right, now, since F is the net force in the direction of the displacement, write it down and plug it into the equation.
 
  • #7
48.1 = (4.2)(9.8) sin theta - .75 (4.2)(9.8) cos theta

Probably a newbie question but what's the best way to solve for theta, from here?
 
  • #8
Stingarov said:
48.1 = (4.2)(9.8) sin theta - .75 (4.2)(9.8) cos theta

Probably a newbie question but what's the best way to solve for theta, from here?

Total work = (F*cos(theta) - Ffr) * displacement.

F is the pushing force, and Ffr is the force of kinetic friction, which equals Ffr = uk * N. The normal force N equals the weight of the chair. Now plug in and solve for theta.
 
  • #9
457.2 = F cos theta - Ffr
457.2 = x * 9.49 -> 457.2 /9.49

48.18 = Fnet
48.18 = F cos theta - 30.87?

Wow if I'm really messing this up let me know, things were clicking until this one and it seems like it should be easy. I understand where I used sin wrong, mistake.
 
  • #10
Stingarov said:
457.2 = F cos theta - Ffr
457.2 = x * 9.49 -> 457.2 /9.49

48.18 = Fnet
48.18 = F cos theta - 30.87?

Wow if I'm really messing this up let me know, things were clicking until this one and it seems like it should be easy. I understand where I used sin wrong, mistake.

There's something more to it, which I forgot to mention in my previous post (excuse me, please). The normal force equals the sum of the chair's weight and the vertical component of the pushing force F, so N = mg + Fsin(theta).
 
  • #11
Fnet = 48.18

48.18 = F cos theta - {(4.2)(9.8) + F sin theta}? Or am I misinterpreting you.


I'm not taking physics this semester but rather next semester, but would like to get as far ahead as possible. Forgive my simple questions eheh.
 
  • #12
Stingarov said:
Fnet = 48.18

48.18 = F cos theta - {(4.2)(9.8) + F sin theta}? Or am I misinterpreting you.


I'm not taking physics this semester but rather next semester, but would like to get as far ahead as possible. Forgive my simple questions eheh.

OK, let's slow down then.

Since the chair has constant velocity, the sum of all forces must equal zero, hence the sum of all horizontal components must equal zero. So, we have:
[tex]F\cos\theta - F_{fr}=0[/tex]. (1)
Further on, since the sum of all vertical components must equal zero, you have:
[tex]N = mg + F\sin\theta[/tex].
You know that the force of friction is [tex]F_{fr}= \mu_{k}N =[/tex]
[tex]=\mu_{k}(mg + F\sin\theta)[/tex].
After plugging that equation into equation (1), you have:
[tex]F\cos\theta = \mu_{k}(mg + F\sin\theta)[/tex]. (2)
Now, since you know the work of the applied force, [tex]W=F\cos\theta d[/tex], you have [tex]F =\frac{W}{d \cos\theta}[/tex]. Plug F into
equation (2), and you'll get an equation which you'll be able to solve for [tex]\theta[/tex]. I hope this helps.
 
  • #13
48.18 = .75 (mg + FsinTheta)

---->F = W/cosTheta*d

48.18 = 30.87 + .75 * 48.18tanTheta?
 
  • #14
Im pretty much beyond lost on this and have no other examples given to go by, so if someone solved it or one like it as a guide, I wouldn't mind. If that is frowned upon on here, then my apologies.
 
  • #15
You have the equation [tex]F =\frac{W}{d \cos\theta}[/tex] . Plugging this into ragou's eqn (2) gives [tex] \frac{W}{d}= \mu_k (mg+\frac{W}{d}\tan\theta) [/tex] which I think you hav. Solving for theta gives [tex] \tan\theta = \frac{1}{\mu_k}-\frac{mgd}{W} [/tex] Plugging your values in gives the solution.

It's a lot easier to work in algebraic terms, and then plug in the values when you have a final equation. It's a good idea to get used to algebra, as this will assist you later on in your studies.

edit: incidentally, your equation above gives the correct answer.
 
Last edited:
  • #16
I see where I went wrong now. Yeah it is easier completely in algebraic terms, interesting.
 
  • #17
Thanks guys, much appreciated.
 

Related to Help with work, angles with friction

1. What is friction and how does it affect angles with work?

Friction is a force that opposes motion between two surfaces that are in contact. When an object is placed on a surface, the force of friction acts in the opposite direction of the object's motion. This can affect the angles of work by making it more difficult to move an object or change its direction.

2. How does the coefficient of friction impact angles with work?

The coefficient of friction is a measure of the amount of friction between two surfaces. It can impact angles with work by determining how much force is needed to overcome the friction between the two surfaces. A higher coefficient of friction means more force is needed to move an object, while a lower coefficient of friction means less force is needed.

3. How can I reduce the effects of friction on angles with work?

There are several ways to reduce the effects of friction on angles with work. One way is to use lubricants, such as oil or grease, to reduce the friction between two surfaces. Another way is to use materials with lower coefficients of friction, such as Teflon or nylon. Additionally, reducing the weight of the object or changing the angle of the surface can also help reduce friction.

4. What is the relationship between the angle of an inclined plane and the force of friction?

The angle of an inclined plane can affect the force of friction in two ways. First, the steeper the angle, the greater the force of friction will be. This is because the weight of the object is distributed more evenly along the surface, increasing the contact between the two surfaces. Second, the angle can also impact the direction of the force of friction, which can affect the motion of the object on the inclined plane.

5. How does friction affect the efficiency of work?

Friction can have a significant impact on the efficiency of work. This is because friction converts some of the energy used to do work into heat, which is a less useful form of energy. As a result, the more friction there is, the less efficient the work will be. This is why it is important to consider and minimize the effects of friction when performing work or calculating work in a scientific context.

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