I\'m really only interseted in C, F, G, and H. H has a typo, we have the mass of
ID: 1795116 • Letter: I
Question
I'm really only interseted in C, F, G, and H.
H has a typo, we have the mass of B already, it's meant to ask how much weight does A need in order to move block B.
I'm concerned that my resulting answer is too low to meet to the requirements of moving the mass of A.
12 degrees 14.) Given the mass of block B is 250 kg and us - 0.22 what is the minimum mass of block A such that it will pull block B up:? 14a.) Draw a FBD for block A. 14b.) Draw a FBD for block B. 14c.)What is the acceleration constraint? 14d.) Collect the sum of forces for block A. 14e.) Collect the sum of forces for block B. 14f.) Find an expression for fs in terms of mb 14g.) Find an expression for fs max in terms of m 14h.) What is the minimum mass of block B to pull up block A?Explanation / Answer
Note: There seems like typing mistake, the typing mistakes in other parts also, f, g and h....it must be mA instead of mB. Later after solving, it was found that the mistake is very first sentence of the question, its actually mass of A is given and B is asked.Took one hour
c] acceleration of A = acceleration of B
f] We have T = mAg
mAgsin 12 degree = fs + T
mAgsin 12 degree = fs + mBg
fs = mBg - mAgsin 12 degree in up the incline direction.
fs = mBg - 250*9.8* sin 12 degree
fs = 9.8mB - 509
g] fs max = uN = umAg cos 12 degree
fs max = 0.22* cos 12 degree *9.8 mA
fs max= 2.11 mA
h] mAgsin 12 degree = fs + mBg
mA*9.8 sin 12 degree - 2.109 mA = 250*9.8
mA*[2.037-2.109] = 250*9.8
mA = 250*9.8/[2.037-2.109] = -34027 kg, Negative answer of block A means if the given mass 250 kg is of block B, then block A can never move downward. So the typing mistake is in mass of block A shoud be given as 250 kg, and B should be asked.
Corrected : mB g = T= mAg sin theta + umA g cos theta
mB = 250* sin 12 degree + 0.22 *250* cos 12 degree
= 105.77 kg