A 50 kg ice skater is gliding along the ice, heading due north...

Question

A 50 kg ice skater is gliding along the ice, heading due north at 4.0 m/s. The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but μ_k = 0. Suddenly, a wind from the northeast exerts a force of 4.0 N on the skater. What is the minimum value of μ_s that allows her to continue moving straight north?

Accepted Answer

Hello, fellow physicist to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A boy is riding his bicycle moving north at 6.0 m per second. There is wind flowing in the direction 30 degrees west of south, exerting a force of 4.5 newtons. Consider there is no rolling friction but static friction which will save the boy from skidding sideways. Calculate the minimum value of the static friction coefficient that will help move towards the north. Consider the combined mass of the boy in the bicycle is 35 kg. Awesome. So we're asked to solve for the minimum value for the static friction coefficient that will help move the boy towards the north direction. And we need to also consider that the combined mass of the boy and his bicycle is 35 kg. So our end goal ultimately is we're trying to figure out what the value for the static friction coefficient is the minimum value rather. So let's look at our multiple choice answers to see what our possible static coefficient value might be A is 0.007. B is 0.008 C is 0.005. And finally D is 0.009. OK. So first off, let us recall and use the particle model, let us consider the bicycle rider to be a particle. And with that in mind, let us create a particle model and a friction model to help us better visualize this problem. So I've gone ahead and devised a diagram here to help us better visualize what is going on. So as we could see we have on our left, we have our particle model. And then on the right, we have our friction model. So as we could see on the right, so let's focus on the right side here. So we have our particle which is represented by a black dot below on the bottom. So Y zero equals 0 m and V zero equals 6.0 m per second. And then it will accelerate upwards going up at delta R vector or vector R. So delta vector R and then at the after point Y one and V one will be some value and it's heading in the north direction. And then we have diagonal black lines here representing the wind at a direction of 30 degrees west of south. And then looking at our friction model here, we have our Y axis is going in the vertical direction and our X axis is in the horizontal direction. And we have our vector FX which is our static friction in the X direction. And then we have our force of the wind which is 30 degrees south of west. Awesome. So the acceleration in the X direction will be zero since the bicycle is prevented from skidding sideways, which we can write as F net of X visits in the X direction is equal to zero newtons. So as we said before FS is the static friction. And F net is the net force and F wind of course is the force of the wind. So we can write that FS minus F wind which I'm just gonna denote as FW to keep things simple. So the force of the wind multiplied by cosine of 60 degrees is equal to zero newtons. So now we need to isolate and solve for FS. So when we do that, we will get that FS is equal to F wind multiplied by the cosine of 60 degrees, which is equal to when we plug in our known variables, 4.5 newtons multiplied by the cosine of 60 degrees which is equal to 2.25 newtons. So note that the maximum value of the static friction will be that FS is less than or equal to the maximum static friction which is equal to the static. So the static friction coefficient, which we can denote as mu subscript S multiplied by M multiplied by G where once again, mu subscript S mu S is the coefficient of static friction and M equals the mass and G equals the gravity. So in order to prevent the bicycle from slipping, us must be greater than or equal to the static friction divided by the mass multiplied by the gravity which is equal to plugging in our known variables. 2.25 newtons divided by 35 kg multiplied by 9.8 m per second squared, which is the numerical value for gravity, which is equal to let me plug that into a calculator 0.0065 which will round 21 significant figure will be 0.007. And that is our final answer. That is our static friction coefficient minimum value. Awesome. So that is when we round the three decimal places. Therefore, the minimum value of the static coefficient or static friction coefficient has to be 0.007. And this corresponds with multiple choice answer letter A which is once again 0.007. Thank you so much for watching. Hopefully that helped. And I can't wait to see you in the next video. Bye.

Key Concepts

Friction

Force and Motion

Equilibrium

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