A rope is tied to a large crate, which is sitting on a flat surface. The coefficient of static friction between the crate and the ground is 0.9. If a person is to pull on the rope with the minimum force needed such that the crate begins to slide, the angle between the rope and the ground should be A) greater than 0 degrees but less than 90 degrees B) 0 degrees (rope is horizontal) C) 90 degrees I know the answer is A.

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Now, I was trying to prove it: horizontal pulling: F=ukmg. greater than 0 degrees but less than 90 degrees: Horizontal component: uk(mg-sinthetaF) (Let F be magnitude of force.) Vertical component: costhetaF Total magnitude: (Pythagoras) and (sin^2theta+cos^2theta=1) 0.81mg-1.8sinthetaF+F^2=Magnitude. How do I prove 0.81mg-1.8sinthetaF+F^2 > 0.9mg?

Hi! I want to follow this from the beginning. You are using two different \(F\)'s, right? So let's say \(F_f=\mu_kmg\) and \(F\) is the minimal force we're looking for. Then we'll define \(\theta\) to be the angle between the ground and rope length, like you did I think.. Then the vertical component is \(\mu_kF_N=\mu_k(mg-F\sin\theta)\) And the horizontal is \(F\cos\theta\) Besides switching the horizontal and vertical components, we're on the same page. And then you look to identify the total force. \(\sqrt{\left(\mu_k(mg-F\sin\theta)\right)^2+\left(F\cos\theta\right)^2~~}\) \(=\sqrt{\mu_k^2(mg-F\sin\theta)^2+\left(F\cos\theta\right)^2~~}\) \(=\sqrt{\mu_k^2(m^2g^2-mgF\sin\theta+F^2\sin^2\theta)+\left(F\cos\theta\right)^2~~}\) \(=\sqrt{\mu_k^2(m^2g^2-mgF\sin\theta+F^2\sin^2\theta)+F^2\cos^2\theta~~}\) \(=\sqrt{\mu_k^2m^2g^2-\mu_k^2mgF\sin\theta+\mu_k^2F^2\sin^2\theta+F^2\cos^2\theta~~}\) And we can't isolate the \(\sin^2\theta+\cos^2\theta\). Also, I see different powers of \(F\) in there.

Although I guess we should use \(\mu_s\) for static friction.

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