resolving forces (CIE mechanics 1 )

OP

i don't know resisting dforces , two type of forces are unknown

Reply 3

Original post by Deffect

i don't know resisting dforces , two type of forces are unknown

Here's an overhead view of the situation.

I'd start by resolving forces perpendicular to the line of motion.

Reply 4

OP

only horizontal motion in this question , ( tugboats can't rise it )

Reply 5

Original post by Deffect

only horizontal motion in this question , ( tugboats can't rise it )

Either you're trolling, or you have a problem with english.

It's an overhead view, i.e. as seen from above. Everything is taking place at ground level.

If you look at your own diagram, you have weight going down, so your "tugs" are flying!

Reply 6

OP

my native language isn't English i'm in 9th grade in my country (English language is very weaker in my country ) ( i don't study igcse , in this year i started with math A level , this a first time in study mechanics)
i think upthrust force can ignored because a heavy barge ( but when i read it again ( in straight road )) so this is problem

(edited 7 years ago)

Reply 7

Original post by Deffect

my native language isn't English i'm in 9th grade in my country (English language is very weaker in my country )

I suggest adding that to your profile so people are aware of it - currently there is nothing there.

( i don't study igcse , in this year i started with math A level , this a first time in study mechanics)
i think upthrust force can ignored because a heavy barge ( but when i read it again ( in straight road )) so this is problem

You need to look at the barge from above, as if in an aeroplane, or from the point of view of a bird flying overhead.

The barge is travelling in a straight line. The two tugs are pulling at an angle, one to the right, and one to the left.

The weight does not come into it. Although the mass does when considering the acceleration along the line of movement.

(edited 7 years ago)

Reply 8

OP

because in straight line so up thrust in one of tugboat will cancel by second boat
so 1800sin(20)/1650=sin(X) so x= 21.9077 degree
acceleration is 0.6 ms^-2 so 0.6*4*1000=1800cos (20)+1650cos(21.9077)-J, j=822.2935N
is correct answer? thanks

Reply 9