Question

hey can someone on here explain how i can find the tension in an object (a rope, in this case)?

basically box a (3kg) is accelerating +1m/s to the right (being pulled by a rope attached to a pulley system, there is no incline) while on the other side of the rope is box m (unknown mass). I know the normal force and the gravitational force cancel out, and that the tension is the same throughout the whole rope, but how would i calculate it? thanks.. (im sorry i accidentally closed the question i just posted this same thing)

Answer 1

this is a diagram of it!

Answer 2

okey! used the formula T(tension)= Fg + (ma)
Fg= 9.8 * m = 29.4, the mass of the box is 3

T= 29.4 + (3*1) = 29.4 + 3 = 32.4 !

so how would i find the value of the box that is being lowered by the rope?

Answer 3

so im using the same equation; T= Fg + (ma)
T= 32.4
m, now equals, the mass of box m
Fg= the gravitational force which is m*9.8
so
32.4=(m*9.8)+(m*1)
32.4= 9.8m + 1m
32.4= 10m
32.4/10= m
3.24 = m
! et viola im not sure if this is the correct way to do this ill tell y'all tomorrow if i got it right owo

Answer 4

always draw a FBD - free body diagram

the easiest FBD to draw would be for the hanging mass


The balance of forces would be \(\sum_{}^{}F_y= Ma_y=T-Mg\)

Answer 5

now, if you look at the sliding block
drawing a FBD
assuming it is a frictionless surface

now looking at the balance of forces in the x direction
\[\sum F_x=m_s a_s = T\]
where \( m_s\)= 3 kg and \(a_s\) = 1m/s^2