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Just checking to see if I've made any mistakes in my hw problems...
1. A block of mass m is at rest on an inclined plane as shown in the figure. Find an expression for the static friction in terms of the mass , the acceleration of gravity, and and the angle shown.
1. The attempt at a solution:
http://www.physicsforums.com/attachm...3&d=1194247920
Newtons 1st Law
sum of forces = 0
mg cos(a) -μs * mg sin (a) = 0
μs = cos(a)/sin(a) = cot(a)
thus: static frictoin = cot(a)*cos(a)
2. A proud new BMW owner drivers her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and the tires are 0.70 (static) and 0.40(kinetic). What is the minimum radius of curvature for the corner in order for the car not to skid?
2. Speed(v) of honda = 25 m/s
Coefficient of kinetic friction(μk) = 0.40

2007-11-05 03:08:02 · 4 answers · asked by korr 2 in Science & Mathematics Physics

Now formula for calculating the value of the minimum radius of curvature for the corner in order for the car not to skid is
r = v^2 / (coeff. of static kin. fric.)*gravity
thus, it's (25 m/s)^2 / 0.40 * 9.81 m/s^2 = 159.2 m

3. A 44.5-N weight is hung on a spring scale, and the scale is hung on a string. The string is lowered at a rate such that the entire assembly has downward acceleration of 4.90 m/s^2. What does the scale read?

3. sum of the forces = m*a

sum of Fy = T - w = ma

thus: (44.5 kg*m/s^2) * ((9.81 m/s^2 - 4.90 m/s^2) / 9.81 m/s^2) = 22.3 N

2007-11-05 03:08:24 · update #1

4 answers

1) friction force = mgsin(a)
2) friction force f = mv^2/R
f< μsmg ( note no skid, so use static friction)
so mv^2/R < μsmg
R > v^2/μsg
3) W - T = ma = Wa/g
T = W(1-a/g)

2007-11-05 13:32:48 · answer #1 · answered by zsm28 5 · 0 0

I can't retrieve the picture.

Based on your clarification I understand that you are looking for an equation governing a car going around a banked by angle A turn of radius R.

We have 3 forces; gravity(W), centripital force(Fc) and force of friction f.

The forces acting along the bank are

Wsin(A) - Fc cos(A) + μWcos(A) =0 or
mgsin(A) - m(V^2/R) + μmg cos(A)=0

R(min)=mV^2/mg(sin(A)+ μ cos(A))=
R(min)=V^2/[g(sin(A)+ μ cos(A))]
if A=0
R(min)=V^2/[g(μ )] =
R(min)=(25)^2/[9.81 x 0.70)=
R(min) = 91m

and the solution to #3 seems to be just fine.

2007-11-05 04:13:49 · answer #2 · answered by Edward 7 · 0 0

IF I have been you, i'd pass decrease back to the SET of theories positioned forth by way of Newton and look at greater ideal...the belief does not state this in any way, shape, or variety. To generalize the theories, "what is going up would desire to come down.

2016-11-10 08:19:13 · answer #3 · answered by Anonymous · 0 0

no idea what you are trying to ask

2007-11-05 04:38:47 · answer #4 · answered by Anonymous · 0 0

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