If a block is at rest on the incline plane at an angle of 30 degrees, would the maximum frictional force that would allow the block to stay in place be calculated using 30 degrees, OR would we have to find the minimum angle of the incline plane needed for the block to slide before we can determine the maximum frictional force? That is, find that minimum angle using tangent inverse of the coefficient of friction given (between the surfaces) and then substitute it in the formula for fs(max) = UsMgcos (minimum angle needed for block to slide) to get the maximum frictional force that would allow the block to stay in place?
122995
A block of mass 90 kg sits on a slope with an angle of inclination 12° above the horizontal. The coefficients of friction between the block and the slope are µk = 0.03 and µs = 0.2.  a.  What is the maximum frictional force that will allow the block to stay in place?  b.  At what angle of inclination will the block start to slide down the slope?  My Question: I am not looking for these two questions to be answered, however, the question has caused me to doubt my understanding of static and kinetic friction re: incline planes.  When i calculated the  the angle at which the block would begin to slide, using the arc tangent of 0.2 i got an angle less than 12 degrees. However, the question is suggesting that the block is at rest with when the incline plane is at 12 degrees. What is my conceptual error in thinking?  My next question:  Wouldnt the minimum angle of the incline plane (tan inverse of 0.2) be needed in answering the first question? that is, having it placed in the formula for the static friction formula = 0.2 x 90kg x 9.81 x cos of (tan inverse of 0.2)? assuming that tan inverse of 0.2 was found already? Is there something wrong with the question or how am i thinking is conceptual flawed?  My Final Question: Is it that the maximum frictional force that allow the block to stay in place is based on whatever angle the incline plane is set at? So, for example, if the incline plane is lowered to an angle of 5 degrees, the max friction force will be adjusted, even though the block would slide at 11.31 degrees? Is it that if lowered to 5 degrees, that the max friction force to overcome would be different?  In other words: If a block is at rest on the incline plane  at an angle of 30 degrees, would the maximum frictional force that would allow the block to stay in place be calculated using 30 degrees, OR would we have to find the minimum angle of the incline plane needed for the block to slide before we can determine the maximum frictional force? That is, find that minimum angle and then substitute it in the formula for  fs(max) = UsMgcos (minimum angle needed for block to slide)?
70741
Is  really independent of area? The friction force, $f_s = \mu_s N$. The equation says that friction only depends on the normal force, which is $ N = W = mg$, and nature of sliding surface, due to $\mu_S$.   Now, less inflated tires experiences more friction compared to well inflated tire. Can someone give clear explanation, why friction does not depend on area, as the textbooks says?
82301
Intro:  In completing , he presents an experiment (go to 42:44) which leaves me baffled.    Experiment:  (I recommend watching the video; see link above.)   There is a $2$ kg block with 2 identical strings attached to it: one at the top, the other at the bottom. The top string is attached to a "ceiling", and the bottom to a "floor". Professor Lewin "stretches" the system (by pulling on the bottom string) with the block not accelerating. One string snaps.     Prediction:   Initially, the top string has a tension of approximately $20$ N, to counter the force of gravity. The bottom string has no tension at all. Then, when Lewin pulls the bottom string, it gains some tension $n$ N. To counter act the force exerted by the bottom string, the top string exerts now $20 + n$ N. I assume that the string with more force will give out sooner, leading me to conclude that the top string will break.     Results:  (This was conducted by Lewin, not me; see link above.)   Trial 1: Bottom string breaks. Trial 2: Top string breaks. Trial 3: Bottom string breaks.     Additional Notes:  The results don't seem consistent. If I was right, I'd expect all 3 experiments to be right; conversely, if I was wrong, I'd expect all 3 experiments wrong, with one exception: the results are more-less random and one result isn't preferred over the other.    Question:        Why was my prediction incorrect?   Was there a flaw in my logic?   Why were the results inconsistent?
70984
I know there are a lot of questions about rolling friction, but I couldn't find any that addressed my specific problem.  Note the following picture where the wheel rotates clockwise. The rolling friction force $\vec{F}_R$ points in the same direction as the rotation.   Normally, rolling friction inhibits the movement (speed $\vec{v}$) of the system. However, since the force is pointing in the "backward" direction, it looks like it is increasing the speed of the wheel. Therefore, it should accelerate the system. This seems unintuitive to me. Can someone please clarify this contradiction?   Thanks!    Update:  Based on the comments and answers, I have reconsidered the diagram and terminology. I mean rolling resistance and no static friction. The deceleration force caused by deformation (and other reasons). In the diagram above, this force is divided into its x- ($\vec{F}_r$) and y- ($\vec{N}$) components.  $\vec{v}$ is not independent of $\omega$ for pure rolling, since $v=R\cdot\omega$. To me, it looks like the resistance $\vec{F}_R$ that slows down the system accelerates $\omega$. And because of the kinematic connection between $\omega$ and $\vec{v}$, the latter must also increase. That's illogical, but I don't understand the reason. Maybe some comments or answers have already addressed that, but I don't understand :( This is really confusing.
88036
I have already read        and many others   But I still have the following problem:   On one hand I've been told that if pure rotation takes place then the object goes on forever. This must imply no frictional force in any form acting on the body. On the second hand I've been told that during pure rotation(no external forces like wind) on a plane surface friction acts opposite to velocity of center of mass (accompanied by change of normal) and slows it down.     Both seem logical enough: The point of contact doesn't slip therefore no friction is required; So no retarding force and it must go on. Things don't go on forever; Friction must be present to slow it down.    So, which hand is correct? Or is there a third hand altogether?
45829
Work can be calculated by $W = Fd\cos\theta$ where $\theta$ is the angle between the force $F$ and the displacement $d$  Let's say there's a ball and a (physically ideal - no friction etc.) robotic arm situated in otherwise empty space. The arm takes the ball, moves it around in a circle, and then returns it exactly to where it started. The arm also returns to its starting position.  In this case, there is no displacement overall. The ball and the robot arm are in the exact same positions as when they began.  Thus, $W = F\cdot 0 \cdot \cos\theta = 0$ So no energy was required to move the ball in a circle.  However, this disagrees with my intuition, because if I were to make a robot arm that would do this, I feel like I would need to give it an energy source (for example, a battery), and that by the time the robot was done, I would have lost energy from that battery.  Does the movement require energy? If so, where did I mess up in my calculation? If not, why would my robot need a battery pack or some energy source in order to make this movement? (If it doesn't, please explain.)
70952
I was wondering about this question since I learned about rolling motion in the chapter on rotational mechanics. I was unable to come to a solid conclusion due to the reasons mentioned below.  The following diagram shows a ball on a frictionless inclined plane and the forces acting on it:    The forces acting on the ball are shown in red and are the normal contact force $N$ and the gravitational force of attraction $mg$. I qualitatively determined the torque of these forces about two axes - one passing through the centre of mass of the ball of uniform density, and the other passing through the point of contact of the ball and the inclined plane. Both of these axes are perpendicular to the screen.  When the axis passes through the centre of the ball, the torque exerted by $mg$ is zero as its line of action meets the axis. Further, the torque exerted by $N$ is also zero due to the same reason. There are no other forces. So, net torque about this axis is zero, and this tempts us to conclude the ball slides down the inclined plane.  When the axis passes through the point of contact, the torque exerted by $N$ is zero but the torque exerted by $mg$ is non-zero. This means the ball must roll i.e., it rotates while moving down the inclined plane. This conclusion is contradictory to the former case.  So, what exactly will happen to a ball kept on a frictionless inclined plane - will it slide or roll?   The following diagram is a visual interpretation of my question (if the terms slide and roll confuses the reader) where the red arrow denotes the orientation of the ball:    Image Courtesy: My own work :)    Please Note: The question -  discusses the case of ball rolling on an inclined plane where friction is present. Since the question -  is marked as duplicate of the former, and has no sufficient details, I planned to ask a new question with additional information.
13556
Static friction : Static friction opposes relative motion between two surfaces. The car is moving at some tangential velocity relative to the road. Thus, static friction should act in the opposite direction. Yet it acts down the incline rather than parallel to the motion of the car. If it acts down the incline, it's essentially helping the car maintain the same relative velocity w.r.t the incline rather than opposing this relative motion.   Kinetic friction :  Consider a block placed on a long, rough plank which is moving to the right at some constant speed, and the block is moving to the right, but slower compared to the plank, so relative to the plank, the block is sliding towards the left. Will kinetic friction act to the left to reduce the kinetic energy of the block, or will it act towards the right to oppose the relative motion of the block and plank? If it acts to the right, doesn't that increase the kinetic energy of the block, which is something kinetic friction should not do?
64697
For context, please consider the following problem from an older AP Physics C: Mechanics practice exam:  Since the ball is rolling without slipping, there is static friction between it and the ground such that torque is exterted on it, making it rotate. If I understood correctly, the frictional force f, in this scenario, opposes the movement of the object, and it is zero if it stops rolling and starts slipping, or ceases contact with the surface. Therefore, at point A on the loop f would point downward, and in point B, since it is still rolling, it would point to the right like so:   However, the CollegeBoard answer sheet shows otherwise. It indicates that at point A f points upwards and at point B it does not act on the ball at all:   Does friction still act on the ball at point B? Is the friction on a rotating body opposite to its movement? If so, did CollegeBoard perhaps get the answer sheet wrong?
120593
If I increase the force $F$, only the normal force $N1$ acting on $m1$ would increase which has no component along the plane, ie. along $m1gsin\theta$, so how would applying this force prevent m1 from sliding?  When I view it from the accelerated frame of the wedge,   It begins making sense. How is this possible? I'm really confused.
86029
I am trying to calculate the minimum stopping distance of a car once the brakes are applied. I know that $F = ma$, and the braking force is $F = \mu N = \mu m g$, so $$a = \mu g.$$ Next, by applying the kinematics equation $$v_f^2 - v_i^2 = 2 a x$$ I found $$v^2 = 2 a d$$ where $a$ is the deceleration, $v$ is the initial velocity, and $d$ is the stopping distance. Then $$d = \frac{v^2}{2a} = \frac{v^2}{2 \mu g}.$$ Is this reasoning right?
77650
A stone is released from an elevator going up with acceleration of $g/2$. What is the acceleration of the stone just after the release?  The answer is $g$. Shouldn't the stone carry the acceleration of the elevator and be $-g/2$?
44410
A small body of mass $m$ slides down from the top of a hemisphere of radius $r$. There is no friction between the surface of the block and the hemisphere. The height at which the body loses contact with the surface of the sphere is? This is how I understood the problem: First of all, the mass does not lose contact with the surface of the sphere at all, considering it is undergoing centripetal acceleration, where the centripetal force is provided by the component of the mass's weight towards the centre of the hemisphere ($mgcos\theta$, where $\theta$ is the angle between the vector of the mass's weight and its component acting towards the centre). Therefore, since only $mgcos\theta$ is responsible for the centripetal force, I can form a relationship like this:  $mgcos\theta\ =\ \large \frac{mv^2}{r}$ $ v\ =\ \sqrt{rgcos\theta}$ Taking 'h' as the height of the mass from the base of the hemisphere. $cos\theta\ =\ \large \frac{h}{r}$ Then the velocity of the mass becomes: $v\ =\ \sqrt{gh}$ The component of the mass's weight along the centre disappears only when $\theta$ becomes $90$ degrees. At this point, it leaves the surface of the hemisphere. Now, the energy of the mass at the topmost point is: $P.E\ =\ mgr$  As the body slides over the hemisphere's surface, it has a tangential velocity given by the expression I had just previously derived. So by the conservation of total mechanical energy of the body, its energy at any other point on the hemisphere is: $T.E\ =\ mgh + \frac{1}{2}mv^2$ $P.E\ =\ T.E$ $gr\ =\ gh + \frac{v^2}{2}$ $gr\ =\ gh + \frac{gh}{2}$ $h\ =\ \frac{2}{3}r$ But this would mean that the body does indeed leave the surface of the hemisphere. It just doesn't add up. Can someone please explain if my approach and assumptions are valid and how I got to this completely contradictory answer?
97949
In a  horizontal surface, a block (cube) is sliding due to a sudden push. When the block slides, there is frictional force which is acting on the block.     Frictional force will have a torque around the center of mass, so why does the block not rotate/roll around (a horizontal axis through) the center of mass instead of sliding?
115616
I have box 2 stacked on box 1. Box 1 is on a frictionless table and accelerated by a force. However, there is friction between box 1 and box 2. Now, I was told that this is the diagram for that problem:  I thought that since m1 and m2 are moving to the right, then the frictional force between box 1 and 2 will be to the left, opposite to the direction of motion. Can someone confirm what is the actual direction of friction for the frictional force between box 1 and box 2? Also, is the action-reaction pair of forces between box 1 and box 2 the normal force of box 1 and the negative normal force of box 1? It is the same issue with another problem I saw online:  In one of the free-body-diagrams, the friction seems to be going to the right (which makes sense since the movement is to the left), but in the lower free-body diagram the friction appears to be going to the right (in the same direction of movement). How is this possible?
77058
A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.   Is my solution wrong? How come I can't get the correct answer?  x=horizontal axis , y=vertical axis Area=XY/2, when X=3,  A=6m^2, dx/dt=0.4m/s  Y=2A/X  dy/dx=-2A/x^2.....A=6,x=3      =-12/9  dy/dt=dy/dx * dx/dt      =-12/9 * 0.4      =-0.533333
53152
In the image below,    All the blocks are frictionless &amp; identical with side of unit length, height $h$, weight $w$ &amp; center of gravity at their geometric centers. The 2 lowest blocks are on solid ground. The distance from the corner of each block to the midpoint of the bottom side of the box above it is given (namely $a1,a2,b1,b2$). forces $F_{ac}$ and $F_{bc}$ are the resultant reaction forces exerted by blocks $A$ &amp; $B$ on $C$.      I am interested in the behavior of these blocks immediately after setting them in this configuration and releasing them, or more specifically: For what relation between $a1,a2,b1$ &amp; $b2$:    do $A,B$ and $C$ move?- case 1 do only $A$ and $C$ move?- case 2 do only $B$ and $C$ move?- case 3 is the configuration stable (does not change at all once set under this condition and then left.)?- case 4    For those interested, here's my approach &amp; what (I think) I know already:  In an attempt to find the limiting conditions (the borderline conditions between equilibrium and non equilibrium), I assumed that initially $C$ will tend to be in equilibrium.(I have no rigorous justification for this assumption, just a hunch that "this isn't where the trouble is").Under this condition, $F_{ac}$ &amp; $F_{bc}$ can be calculated, and the moments due to their "equal and opposites" ($F_{ca}$ &amp; $F_{cb}$) about $P1$ and $P2$ can be obtained as: $$ M_{ca}(x,y)=w(.5+a1-a2-x)/(x/y+1)$$ $$ M_{cb}(y,x)=w(.5+b1-b2-y)/(y/x+1) $$ where $x$ &amp; $y$ are the perpendicular distances of the respective forces from the center of $C$.   With some ad-hoc and shaky logic here's what I arrived at:   when $$w\cdot a_2 &lt;M_{ca}(.5,.5)\quad \&amp; \quad w\cdot b_2 &lt;M_{cb}(.5,.5)$$ case-1 occurs with $A$ &amp; $B$ touching $C$ only through it's vertices. when $$w\cdot a_2 =M_{ca}(x1,.5)\quad \&amp; \quad w\cdot b_2 &lt;M_{cb}(.5,x1)$$ where $a1\leqslant x1&lt; .5$ case-1 occurs with $C$ rotating with $A$ maintaining a surface of contact with $A$ but only a point contact with $B$, but if $$\boldsymbol{w\cdot a_2 =M_{ca}(x1,.5)\quad \&amp; \quad w\cdot b_2 \geqslant M_{cb}(.5,x1)}$$ the configuration is stable (case-4). when $$w\cdot a_2 &gt;M_{ca}(a1,.5)\quad \&amp; \quad w\cdot b_2 &lt;M_{cb}(.5,a1)$$ case-3 occurs, but if $$\boldsymbol{w\cdot a_2 &gt;M_{ca}(a1,.5)\quad \&amp; \quad w\cdot b_2 &lt;M_{cb}(.5,a1)}$$ the configuration is stable (case-4). The above conditions with $a$ &amp; $b$ exchanged along with their corresponding variables.   But I have no way to check this or provide a satisfactory argument for these conditions, especially the parts written in bold (I arrived at it by putting various combinations of arguments for $M_{ca}$ &amp; $M_{cb}$ &amp; thinking about what would happen in each case). Is this set of conditions right?. What would be a good approach with a logical progression of steps to solve it?