for part a i subtract t-150n=15kg*a because i dont want tension right? solving f

Question

for part a i subtract t-150n=15kg*a because i dont want tension right? solving for a?

also, reason why t-600sin40= -60kg*a is f=ma? so -60kg*a ??

for part b i dont quite get it..

t=15kg*3.13m/s^2 was f=ma i guess but why do you have to add 150N ?

t=mg*a*m ? why?

dont get it.

Let M = 60 kg, m = 15 kg and the angle with the horizontal be 40 Let the string and the pulley be massless and there's no friction in the pulley. Draw free-body diagrams for M and m. Calculate the acceleration of M and m. Calculate the tension in the string connecting them.

Explanation / Answer

for part (a)

from the free body diagram of block of mass M

net force = ma (F = ma)

net force = W sin 40 -T

so

T-Wsin40 =- ma

T-600sin40 = -60a........................(1)

from the free body diagram of block of mass m

net force = ma

net fore = T - 150

T-150 = ma (F= ma)

T - 150 = 15a............(2)

substracting (2) from (1)

T- 600sin40 -T +150 = -60a - 15a

-385 + 150 = -75a

-75a = -235   

75 is weight so it hs dimension kg

and dimensions of 235 is kgm/s2

so kg - kg will cancel out

so a will have dimension m/s2

hence a = 3.13 m/s2

for part (b)

from the free body diagram of block of mass m

T - 150 = 15a   (F=ma) , where F is net force and hre net force = T - 150

SO

T = (15 * 3.13 ) + 150    .....................(a = 3.13 m/s2 )

T = 46.95 + 150 = 196.95 N

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