Question

6. A hiker with a mass of 85 kg wants to drag a 200 kg sled over snow. She does this by tying a rope to the sled (at ground level) and running it over her shoulder. The rope running between her and the sled makes a 45 degree angle with the horizontal If the coefficient of friction between the sled and the snow is 0.1, what must the coefficient of friction between her tbots and the ground be for her to move the sled? Note: The algebra for this problem can get messy. If you write down a system of four equations and identify the four unknowns you want to solve for, you will get most of the credit for this problem. This is the messiest algebra you will see this semester.

Answer 1

F.BD m = 85kg I ng m= 200kg Refer the F. B.D of sled and hiker Nit Tsin 45o = mg o Forsted T Coss5° >fi © fa > T cos 45o - Ne = mg + Tsintsa @ For likes 7 M₂ Na>, I coses & T 20545 > HN → H2 (mg + tsinese) > I coslogo > lng-Tsin450) $ TC coses + fly sin ese) > Hi mg * Talling | Mamin 2 52 xVZ Tos45 mg +1 Sindeso 85*9.8 + 252 Coses + Misinus! = 0.176 To just move the sled 178.2 — 833 +17872 ~ 0.18 T= 012 DOX 9.8 ring Ha 152 + y sexul - Ans 1 Cos445 +14, Sinless = 252N