Mechanics help?

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Reply 20

CheeseIsVeg

Original post by petrus123

OK, thanks! I'll give it a try.

Awesome, good luck, see what you get and compare it to that of @the bear

(edited 7 years ago)

Reply 21

peterw55

Original post by CheeseIsVeg

Awesome, good luck, see what you get and compare it to that of @thebear

OK, I'm just slightly confused about why @thebear has his contact force arrow pointing upwards (it says down on the lift), although I did that at first too, so maybe I missed something.

Reply 22

CheeseIsVeg

Original post by petrus123

OK, I'm just slightly confused about why @thebear has his contact force arrow pointing upwards (it says down on the lift), although I did that at first too, so maybe I missed something.

Be summoned @the bear

Reply 23

peterw55

I know this should be ridiculously simple, but I'm not quite sure how to solve it. Could anyone help with the last bit?

Reply 24

the bear

Original post by petrus123

OK, I'm just slightly confused about why @thebear has his contact force arrow pointing upwards (it says down on the lift), although I did that at first too, so maybe I missed something.

the lift is pressing up against the passenger's feet. *

Reply 25

peterw55

Original post by the bear

the lift is pressing up against the passenger's feet. *

Yes, but how do we know that that force is 294N?

Reply 26

the bear

Original post by petrus123

Yes, but how do we know that that force is 294N?

she is not sinking into the floor or lifting up from the floor. with respect to the floor she is in equilibrium. *

Reply 27

peterw55

Original post by the bear

she is not sinking into the floor or lifting up from the floor. with respect to the floor she is in equilibrium. *

OK, thanks. But what about the 294N downward contact force? Sorry if I'm missing something.

Reply 28

the bear

Original post by petrus123

OK, thanks. But what about the 294N downward contact force? Sorry if I'm missing something.

since she is in equilibrium with respect to the floor the upwards and downwards forces must be equal. *

Reply 29

peterw55

Original post by the bear

since she is in equilibrium with respect to the floor the upwards and downwards forces must be equal. *

So you could then use 9.8(40+m)-T-294=3(40+m)?
If so, how do you find T?

Reply 30

peterw55

Original post by petrus123

So you could then use 9.8(40+m)-T-294=3(40+m)?
If so, how do you find T?

Anyone?

Reply 31

the bear

Original post by petrus123

So you could then use 9.8(40+m)-T-294=3(40+m)?
If so, how do you find T?

if you mean T to be the force in the string then it is not relevant as there are two equal and opposite tensions in the string which cancel out. *

Reply 32

peterw55

Original post by the bear

if you mean T to be the force in the string then it is not relevant as there are two equal and opposite tensions in the string which cancel out. *

OK, thanks!

Reply 33

peterw55

Original post by petrus123

So you could then use 9.8(40+m)-T-294=3(40+m)?
If so, how do you find T?

I solved the problem a while ago, but I came back to this because I'm curious as to why the 294N force is only included in the upwards direction in the diagram if it is actually in both directions (so that the girl is in equilibrium relative to the lift floor). Sorry if this is a stupid question and/or obvious.