# Help with work, angles with friction

1. Dec 8, 2006

### Stingarov

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.

2. Dec 8, 2006

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. Dec 8, 2006

### Stingarov

F friction (Uk * Fn) and Applied Force, correct?

4. Dec 8, 2006

Correct. Now, what is your next step?

5. Dec 8, 2006

### Stingarov

W = F * d,

457.2 = F * 9.49

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

6. Dec 8, 2006

Right, now, since F is the net force in the direction of the displacement, write it down and plug it into the equation.

7. Dec 8, 2006

### Stingarov

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. Dec 8, 2006

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. Dec 8, 2006

### Stingarov

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. Dec 8, 2006

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. Dec 8, 2006

### Stingarov

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. Dec 8, 2006

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:
$$F\cos\theta - F_{fr}=0$$. (1)
Further on, since the sum of all vertical components must equal zero, you have:
$$N = mg + F\sin\theta$$.
You know that the force of friction is $$F_{fr}= \mu_{k}N =$$
$$=\mu_{k}(mg + F\sin\theta)$$.
After plugging that equation into equation (1), you have:
$$F\cos\theta = \mu_{k}(mg + F\sin\theta)$$. (2)
Now, since you know the work of the applied force, $$W=F\cos\theta d$$, you have $$F =\frac{W}{d \cos\theta}$$. Plug F into
equation (2), and you'll get an equation which you'll be able to solve for $$\theta$$. I hope this helps.

13. Dec 8, 2006

### Stingarov

48.18 = .75 (mg + FsinTheta)

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

48.18 = 30.87 + .75 * 48.18tanTheta?

14. Dec 8, 2006

### Stingarov

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. Dec 8, 2006

### cristo

Staff Emeritus
You have the equation $$F =\frac{W}{d \cos\theta}$$ . Plugging this into ragou's eqn (2) gives $$\frac{W}{d}= \mu_k (mg+\frac{W}{d}\tan\theta)$$ which I think you hav. Solving for theta gives $$\tan\theta = \frac{1}{\mu_k}-\frac{mgd}{W}$$ 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.

Last edited: Dec 8, 2006
16. Dec 8, 2006

### Stingarov

I see where I went wrong now. Yeah it is easier completely in algebraic terms, interesting.

17. Dec 8, 2006

### Stingarov

Thanks guys, much appreciated.