A 3.0-m-long ladder, as shown in Figure 12.35, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.40. What is the minimum angle the ladder can make with the floor without slipping?

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Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. An extendable paw that is 4 m in length is placed against a smooth vertical surface. The static friction coefficient between the pole and the ground is 0.50 calculate the smallest angle the pole can form with the ground to stay in position without sliding. So that is our angle is to calculate the smallest angle the pole can form with the ground to stay in position without sliding. OK. So we're given some multiple choice answers for theta to represent the smallest angle and they're all in degrees. So let's read them off to see what our final answer might be. A is 55. B is 90 C is 30 D is 45. OK. So let's start by drawing a little diagram to help us understand what's going on in this problem to help us visualize it. So let's do this. OK. So we have our po which is vertical to the ground. Then let's make it into a nice little triangle here. So we can get our angles to help us solve or our smallest angle. So the distance OK. So this is vector N two oh So a pole is extend a pole is placed against a smooth surface. So to be clear, so the pole is like leaning. So it's forming a triangle with the smooth surface. So we can consider the smooth vertical surface to be like a wall or something. OK. So this line right here is L divided by two. So the length divided by two and this right here is vector N one. And then the distance from here to here we'll call it D two. And then for here we have a force of FG is equal to MG. Then we have an angle theta here and we have a force fs in the distance of D one for this little tidbit. OK. So first off, let us note that in order for the pole to stay in its current position, it should be in a translational and rotational equilibrium state. So it has to have transitional and rotational equilibrium in order for the pole to remain as it is. So please recall that translational equilibrium states that the sum of FX is equal to zero. And that, and that the sum of FY is also equal to zero. OK. So, and rotational equilibrium states that the sum of torque is equal to zero. So when we apply translational equilibrium condition to the X axis. We get that the sum of FX is equal to zero. And looking at our free body diagram above, we can state that the sum of FX is equal to N two minus FS. Therefore, we can write that N two minus FS equals zero. And when we solve for N two, N two will be equal to FS. And let's call this equation one. So now when we apply the translational condition to the Y axis, we will get the sum of Fy equals zero. So we can then go on to say when we look at our free body diagram, we can write that N one minus MG equals zero. So like we did before we need to solve for N one. So when we do that, we get N one is equal to MG and this will be equation two. So note that the pole will form an angle, the smallest angle with the ground to stay in position without sliding. When the static friction is at its maximum value when it can be written as. So literally, this is the condition for the, for the smallest angle with the ground, the same position without sliding. So the condition for that to be the case would be FS max. So this is the max is equal to the co effi the static coefficient multiplied by N one. So we can take this value and substitute it into equation one. So we can state that N one or I should say N two, N two is equal to the static, the coefficient of static friction multiplied by N one. OK. So note that equation two states that N one equals MG. So we can write this equation to state that N two equals MS or I guess you'd say the coefficient or the static coefficient multiplied by MG. Let's call this equation three. So now we can apply the condition of rotational equilibrium. Let us start by selecting the pivot point to be at the bottom of the ladder or the bottom of the pole. So that would be the sun of torque is equal to zero to form our or, and then also taking into account from our diagram, the net torque from the bottom of our pole is the sum of torque is equal to D one multiplied by G is yeah, D one multiplied by M multiplied by G minus D two multiplied by N two. So D one multiplied by M multiplied by G minus D two multiplied by N two is equal to zero. So now we can plug in N two from equation three. So D one multiplied by M multiplied by G minus D two multiplied by the co the static coefficient multiplied by M multiplied by G equals zero. So we can go on even further to say that D one minus D two multiplied by the coefficient. A static friction is equal to zero. So D one equals D two multiplied by the coefficient of static friction. And this would be equation four and we can call it equation for. So note that from the diagram, D one equals L capital L divided by two multiplied by cosine of theta. And that D two is equal to L multiplied by sine of the. So now we could substitute are these values into equation for. So capital L divided by two multiplied by cosine of theta is equal to L multiplied by sine, theta multiplied by the coefficient of static friction. So we need to rearrange this equation to solve for theta using a little bit algebra. And when we do that, we get theta is equal to inverse tangent of one half divided by the coefficient of static friction. So let's plug in our value now because we know that our coefficient for static friction in this case is 0.50. So let's plug that in. So tangent. So theta equals tangent or tan inverse of one half divided by 0.50. So when we plug into a calculator, we should get 45 degrees. So the smallest angle, the pole conformed to with the ground to stay in the position without sliding is 45 degrees. Fantastic. We did it. So let's look at our multiple choice sensors to see what the correct answer is. It looks like it's the letter D theta equals 45 degrees. Thank you so much for watching. Hopefully that helps and I can't wait to see in the next video. Bye.

A 3.0-m-long ladder, as shown in Figure 12.35, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.40. What is the minimum angle the ladder can make with the floor without slipping?

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Key Concepts

Static Friction

Equilibrium Conditions

Torque