Can someone explain to me why the answer to #224 and #224 are static frictions? The question is if the car is being towed or if the car is moving under its own power, what is the frictional force between the tire and the surface? The solution says it's static because contacting surfaces aren't sliding with each other. But isn't the tire in contact with the surface?
EK Physics Question
- Thread starter shffl
- Start date
When a wheel is rolling properly, the rubber does not slide over the road surface. It rolls over it. The friction is thus static friction. You only get kinetic friction when the wheel is not properly gripping the road. This is typically when you are skidding out, fishtailing, or generally doing something boneheaded with your car.
- Joined
- Jan 20, 2008
- Messages
- 1,270
- Reaction score
- 19
When a tire is rolling the motion of one surface on the other is a "sticking" force where one part of the tire is lifted off the ground continually by the rotational movement but is at no point sliding against the ground. I hope that helps.
- Joined
- Apr 15, 2006
- Messages
- 3,366
- Reaction score
- 3
It's been a while, so I might have a bit of difficulty explaining, but think of the tire in relation to the ground. The wheel's outer edge travels at the same velocity is the surface of the ground, so a single point of the ground stays connected with a single point on the tire throughout the movement.
This page should help you understand it more.
http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/frictire.html
The far bottom describes how the tire and the road are instantaneously at rest with each other when there is full traction.
This page should help you understand it more.
http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/frictire.html
The far bottom describes how the tire and the road are instantaneously at rest with each other when there is full traction.
Thanks I get it now. I just ran across another problem hopefully someone can help as well. The question is regarding tension of a rope with 10kg mass hanging on it in an elevator that's descending at 3m/s^2. What I gathered was there's upward force due to tension and downward force from the elevator and gravity so the equation would look like T = mg + ma. I found out it was wrong and the correct equation used was T + ma = mg. Can someone explained to me as to why it's like this when ma and T are in opposite directions? The solutions on EK1001 isn't the greatest when it comes to explaining =[
#280 and #281 took me really long to solve and I was wondering if anyone knows an easier and more efficient way of getting to the answer.
(The figure is 2 ropes on a wall (T1 and T2 with theta 1 and theta 2) converging onto T3 where the mass is hung)
If m = 10kg ;theta1 = 60; theta2 = 30; what is the tension on T2? Answer: 50N
If m = 100kg ;theta1 = 30; theta2 = 30; what is the tension on T1? Answer: 1000N
The solution to both problems state that since object is in equilibrium, horizontal and vertical forces are equal and presented me with 2 equations:
T1 cos theta1 = T2 cos theta2
mg = T1 sin theta1 + T2 sin theta2
If I substitute T1 in 2nd equation with T1 = (T2 cos theta2)/cos theta 1 , I get a complicated equation: mg = T2[(0.87^2/0.5) + 0.5] and mg = 2 T2. Since mg = 100N then T2 = 50N for the first problem. Following the same method I got 1000N for 2nd problem.
#280 and #281 took me really long to solve and I was wondering if anyone knows an easier and more efficient way of getting to the answer.
(The figure is 2 ropes on a wall (T1 and T2 with theta 1 and theta 2) converging onto T3 where the mass is hung)
If m = 10kg ;theta1 = 60; theta2 = 30; what is the tension on T2? Answer: 50N
If m = 100kg ;theta1 = 30; theta2 = 30; what is the tension on T1? Answer: 1000N
The solution to both problems state that since object is in equilibrium, horizontal and vertical forces are equal and presented me with 2 equations:
T1 cos theta1 = T2 cos theta2
mg = T1 sin theta1 + T2 sin theta2
If I substitute T1 in 2nd equation with T1 = (T2 cos theta2)/cos theta 1 , I get a complicated equation: mg = T2[(0.87^2/0.5) + 0.5] and mg = 2 T2. Since mg = 100N then T2 = 50N for the first problem. Following the same method I got 1000N for 2nd problem.
Last edited:
- Joined
- Jan 20, 2008
- Messages
- 1,270
- Reaction score
- 19
Haha Luckily, I was just open to the page on the EK forum for that problem.
Here you go:
This problem is based on translational equilibrium, which says that if an object is not accelerating in a particular direction, then the net force on it in that direction is 0. Since the mass is hanging there, it is neither accelerating in the left-right direction or the up-down direction. Similarly, the tiny piece of rope that is located at the intersection of all three pieces of rope is not accelerating either in the up-down or left-right direction.
This means that, from the point of view of that tiny piece of rope right at the intersection, the total forces pulling up exactly cancel the total forces pulling down, and the total forces pulling left exactly cancel out the total forces pulling right.
Rope T3 pulls downward only from that point (its not oriented at all in the left or right direction. And we can calculate that force (tension) in the rope, because that piece of rope is supporting the weight of the block.
Recall the formula for weight is:
m*g
weight of block = 10*10 = 100 N
Therefore, T3 is 100 N (that piece of rope pulls at both its ends with a force of 100N; pulling up on the block to support its weight, and also pulling down on our intersection point).
Rope T1 pulls on our intersection point in two directions, both upwards and to the left. Rope T2 pulls on our intersection point in two directions, both upwards and to the right.
When these ropes pull in both directions, we have to take our total tension and resolve it into its two components, the up/down component 👍, and the left/right component (x)
This is done with trig, using the angles:
The left component of rope T1 is T1*cos(theta1), since cos(theta) = length of adj / length of hypotenuse.
The up component of rope T1 is T1*sin(theta1), since sin(theta) = length of opp / length of hypotenuse.
Similarly, the right component of rope T2 is T2*cos(theta2).
The up component of rope T2 is T2*sin(theta2)
So all the forces acting on our intersection point are:
UP
T2sin(theta2)
T1sin(theta1)
DOWN
T3 = 100 N
LEFT
T1cos(theta1)
RIGHT
T2cos(theta2)
Again, since the intersection point is not accelerating, the up must cancel the down, the left must cancel the right.
We also know that theta1 = 60, theta2 = 30 so:
Total UP = Total DOWN
T2*sin(30) + T1*sin(60) = 100
sin(30) = 0.5, sin(60) = .86
0.5*T2 + 0.86*T1 = 100
AND
Total LEFT = Total RIGHT
T1cos(60) = T2cos(30)
0.5*T1 = 0.86*T2
We want to know T2, so using the second equation we can write:
0.5*T1 = 0.86*T2
T1 = 0.86T2 / 0.5 = 1.7 * T2
And substitute this T1 into our first equation:
0.5*T2 + 0.86*T1 = 100
0.5*T2 + 0.86 * (1.7 * T2) = 100
0.5*T2 + 1.46*T2 = 100
1.96*T2 = 100
T2 = 50 N
Check out the forum for answers to more of those questions. Registration is free.
http://www.examkrackers.com//forum/viewtopic.php?t=15436
