what is a max? Do not round intermediate calculations; however for display purpo

Question

what is a max?

Do not round intermediate calculations; however for display purposes report intermediate steps rounded to four significant figures. Give your final answer(s to four significant figures. A crate of weight W 549 lb has been attached to a pickup truck via a rope whose tensile strength is T 344 lb. If the truck and crate start at rest with 6 270, determine the maximum acceleration of the truck such that the rope does not break. Determine the solution for the case in which friction between the crate and the ground is not negligible and 0.4 and H k 0.28 ft/s

Explanation / Answer

>> Considering Crate, forces acting are :-

1), W = Weight of Crate = 549*2.2 = 17677.8 lb-f

2). T = Tension in Cable , maximum = 344*32.2 = 11076.8 lb-f,

and is acting at angle 27 degree with horizontal

3). f = Friction Force, acting horizontally in backward direction

4). N = Normal Reaction between ground and crate

Let Net Acceleration = a

>> As, along Y-Direction,

N = W - Tsin27

=> N = 17677.8 - 11076.8*sin27

=> N = 12649.04 lb-f

So, fmax = Maximum Friction Value = 0.28*N [ We will Use Kinematic Friction Coefficient, as it is moc=ving with acceleration a]

=> f = 12649.04*0.28 = 3541.731 lb-f

T*cos27 - f = ma

=> 11076.8*cos27 - 3541.731 = 549.a

=> Solving,

=> a = 11.526 ft/s2 .....REQUIRED ACCELERATION...........ANSWER......

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