I just need someone to walk me through these two problems ifthey can... I can br

Question

I just need someone to walk me through these two problems ifthey can... I can break up everything into the components, but thendon't really knwo where to start from there.. I hae both of theanswers, just trying to work to them like i said.. pelase, if youcould help me, that'd be great. thanks A 289-kg log is pulled up a ramp bymeans of a rope that is parallel to the surface of the ramp. Theramp is inclined at 30.0° with respect to the horizontal. Thecoefficient of kinetic friction between the log and the ramp is0.900, and the log has an acceleration of0.700 m/s2. Find the tension inthe rope.
A 230-kg crate rests on a surface that isinclined above the horizontal at an angle of 19.4°. A horizontal force (magnitude =516 N and parallel to the ground, not theincline) is required to start the crate moving down the incline.What is the coefficient of static friction between the crate andthe incline? I just need someone to walk me through these two problems ifthey can... I can break up everything into the components, but thendon't really knwo where to start from there.. I hae both of theanswers, just trying to work to them like i said.. pelase, if youcould help me, that'd be great. thanks A 289-kg log is pulled up a ramp bymeans of a rope that is parallel to the surface of the ramp. Theramp is inclined at 30.0° with respect to the horizontal. Thecoefficient of kinetic friction between the log and the ramp is0.900, and the log has an acceleration of0.700 m/s2. Find the tension inthe rope.
A 230-kg crate rests on a surface that isinclined above the horizontal at an angle of 19.4°. A horizontal force (magnitude =516 N and parallel to the ground, not theincline) is required to start the crate moving down the incline.What is the coefficient of static friction between the crate andthe incline?

Explanation / Answer

probl 2; a=0;              Fx= P cos - fk =max =0         and    fk = FN = 0, sothe expression becomes Horizontal     P cos -kFN =0         (1) Vertical         Fy=Psin + FN - mg = may =0    (2) Solving (2) equation for the FN and substituting into Equation(1), we get P cos - k (mg - P sin ) =0 solving for k = P cos / mg - P sin = (516N) cos 19.4/ (230 kg) (9.8 m/s2) -(516N) sin 19.4 = k = ...0.23

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