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don't understand force exerted by A on B

Hey, I need help with the question. I got the acceleration to be (3/8)G
(edited 4 years ago)

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Reply 1
Original post by dasda
Hey, I need help with the question. I got the acceleration to be (3/8)G


Work out the tension in the rope.

Now consider the block B. It has its own weight acting down. It has the reaction force acting up on it (this is just going to be the same as the tension). It has the force exerted by A acting down on it.

Since B is accelerating at 3/8 g we have that the following must hold due to Newton's 2nd Law:

(Weight) + (Force Exerted by A) - (Tension) = (Net Force on B given its acceleration)
(edited 4 years ago)
Original post by RDKGames
Since B is stationary in the frame of reference of the scale pan, then the vertical forces must balance. I.e.

(Tension) = (Weight) + (Force Exerted by A)


:holmes: Don't think that's corrrect as the frame of reference is accelerating.
Reply 4
Original post by ghostwalker
:holmes: Don't think that's corrrect as the frame of reference is accelerating.

do we use F=ma. Im struggling with these kind of questions
Reply 5
Original post by RDKGames
Work out the tension in the rope.

Now consider the block B. It has its own weight acting down. It has the reaction force acting up on it (this is just going to be the same as the tension). It has the force exerted by A acting down on it.

Since B is stationary in the frame of reference of the scale pan, then the vertical forces must balance. I.e.

(Tension) = (Weight) + (Force Exerted by A)

The mark scheme shows 2g-reaction force of a = 2a
I think it must be in equilibrium because it says 'sits'
or am i being stupid
gimme a min, im trying to solve it :P
Reply 8
no I think its moving
what is the answer?
I dont want to completely embarrass myself before I say my answer :rofl:
Reply 11
12.25
Reply 12
Original post by Serenity-M
I dont want to completely embarrass myself before I say my answer :rofl:

Don't worry. I hate these types of questions
hmmmmmm :hide:
ohhhhh ok. Lemme keep trying :rofl:
resolve the force parallel to the slope, and resolve for the pan and mass.
for the slope its T - component of gravity = 5a, for the mass and pan its 7g - T = 7a
use that to calculate acceleration, then you know that for A, using F=ma, 2g - F = 2a, since you know have A you can solve.
Original post by Serenity-M
what is the answer?

i got 12.25
Original post by ghostwalker
:holmes: Don't think that's corrrect as the frame of reference is accelerating.


Whoops, completely butchered that.

Should be good now.
Reply 18
how would you do part b ii of this question.@RDKGames
Original post by dasda
how would you do part b ii of this question.@RDKGames


This is a direct analogue of the question you've asked first.

Just apply the same ideas.

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