I've been doing some practice force questions and there are a few I don't understand. I have the solutions to them, so my questions will be relatively short, but I just need to have some things cleared up.(adsbygoogle = window.adsbygoogle || []).push({});

The first one:

"A circus performer of weight W is walking along a 'high wire' as shown in the diagram below (first attachment). The tension of the wire is:

a.) Approximately W

b.) Approximately W/2

c.) Much less than W

d.) Much more than W

e.) Depends on whether he stands on one or two feet"

The answer is D and I'm not sure I understand completely why, so I'm just looking for someone to check my thinking here. I think it is because the tension of the rope has an x- and y-component, and in order for the performer to be held up, the y-component must be greater than or equal to W. But that is just the y-component and because the tension also has an x-component, the overall tension has to be greater than W.Is my thinking here correct?

The second:

"A 12 kg crate rests on a horizontal surface and a boy pulls on it with a foce that is 30 degrees above the horizontal. If the coefficient of static friction is μ_{s}=0.52, what is the minimum force he needs to start the crate moving?"

I drew a free body diagram and resolved the forces into x- and y-components to find:

Along x: Fcosθ - F_{f}= ma_{x}

where F is the force by the boy and F_{f}is the friction force

Along y: N - mg + Fsinθ = ma_{y}= 0

where N is the normal force and mg is the weight of the crate.

So N = mg - Fsinθ

F_{fmin move}= μ_{s}*N = μ_{s}*(mg - Fsinθ)

I now have an expression I can substitute for F_{f}in the "Along x" equation. However, in order to solve, I have to set ma_{x}= 0, in order to get:

F_{f}= Fcosθ

So, μ_{s}*(mg - Fsinθ) = Fcosθ

And then I can solve for F. After all that rambling, my question is:how can I say the acceleration along x is 0 if I'm looking for the minimum force that willmovethe crate?

Lastly,

"A stuntman is practicing for a 'death-defying' loop-the-loop stunt on his motorcycle. The track is essentially a circle that is perfectly vertical, and he is riding his motorcycle inside the circle. The circular loop has a radious of 3.0 m. Together, the mass of the stuntman and his motorcycle is 200 kg. If he drives his motorcycle very fast, he is sure to stay on the track at all times while doing the stunt. If he drives too slowly, he will not stay on the track (resulting in a crash).

Calculate the smallest speed necessary for the stuntman and motorcycle to stay on the track."

I've attached the diagram from the solutions (second attachment), showing the Free Body Diagram of the motorcycle at the top and bottom of the track.

The solution says: "The motorcycle just barely stays on the track at the top if N = 0" and then proceeds to solve mg = mv^{2}/r for v.

I was wondering if anybody could help me to understand why:

-The lengths of the force vectors are the way they are (N is shorter than mg at the top, mg is shorter than N at the bottom)

-N = 0 at the top if the motorcycle just barely stays on the track.

Ididunderstand this about a week ago... but it was such a stretch that it kind of slipped my mind. Any help is appreciated!

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# Homework Help: Forces - friction, centripetal

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