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Q: coefficient of static friction ( No Answer,   1 Comment )
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 Subject: coefficient of static friction Category: Science Asked by: mersedeh-ga List Price: \$2.00 Posted: 16 Oct 2005 01:37 PDT Expires: 26 Oct 2005 02:05 PDT Question ID: 580852
 ```The coefficient of static friction between the m = 2.60 kg crate and the 35.0° incline of Figure P4.41 is 0.290. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?```
 ```Whew, this should be easier to imagine! (Just been to the Science-Astronomy page. Just burned me brain imagining them burstars, galactic plane, Myr-periods, Oort Cloud. Zeez, t'was Chaos Cloud, now Oort Cloud. Maybe if I did not escape from that page, I'd encounter Gungbung Clouds, whatever they are.) So Let me imagine that Figure P4.41. (It should be in a book or something in this planet of ours.) Maybe the 2.60 kg crate is not being controlled by a rope being pulled by a pulley or hand. Maybe the crate is left by itself to slide down the incline. And if the crate tends to slide down, a force F, perpendicular to the incline, must be applied on the crate to stop the motion. And you are asking what minimum F is needed. Well, without F yet, only the friction force due to the component perpendicular to the incline of the crate's weight is there to prevent the sliding down of the crate. And the force causing the motion is the component parallel to the incline of the crate's weight. Crate's weight = m*g = (2.6 kg)*(9.8 ft/sec/sec) = 25.48 newtons Weigth's component perpendicular to incline = (25.48)cos(35deg) = 20.87 newtons Available friction force = (0.290)(20.87) = 6.05 newtons. Sliding force = (25.48)sin(35deg) = 14.61 newtons Umm, the sliding force is a lot greater than the friction force fighting it. The crate will slide! A giant forefinger is needed to press the crate gently to prevent the crate from sliding down. 14.61 minus 6.05 = 8.56 newtons <---to be neutralized. friction force = (coefficient of friction)*(perpendicular force) So, 8.56 = (0.290)*(F) F = 8.56 / 0.290 = 29.52 newtons. Therefore, the giant needs to exert only enough force perpendicular to the incline of 29.52 newtons to stay the crate.```