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1. physics homework help

1. A 400 N child and a 300 N child sit on either end of a 2.00 m long seesaw. where along the seesaw should the pivot support be placed to ensure rotational equilibrium?

2. Fk physics. I hate this class.

3. I dunno the exact nnumbers but your answer should be over 1.0 m and under 2.0 m from the 300n child. Im sure that doesnt help what so ever lol

4. Originally Posted by 140wrestlerK
1. A 400 N child and a 300 N child sit on either end of a 2.00 m long seesaw. where along the seesaw should the pivot support be placed to ensure rotational equilibrium?
sum torque clockwise= sum torque ccw
torque=fd

so 400d1=300d2
d1+d2=2
d1=2-d2
400(2-d2)=300d2
800-400d2=300d2
800=700d2
d2=1.14m from the 300N child.

Anyone wanna check my work for me?

5. thankyou so much, repped. that actually makes a lot of sense. i hate how physics makes as much sense as reading a language you dont know until you have it broken down and explained.

if anyone else is willing to explain another problem id really appreciate it.
a bridge 20m long weighing 4x10^5N is supported by two pillars located 3m from each end. if a 1.98x10^4N car is parked 8m from one end of the bridge how much force does each pillar exert?

6. Originally Posted by 140wrestlerK
thankyou so much, repped. that actually makes a lot of sense. i hate how physics makes as much sense as reading a language you dont know until you have it broken down and explained.

if anyone else is willing to explain another problem id really appreciate it.
a bridge 20m long weighing 4x10^5N is supported by two pillars located 3m from each end. if a 1.98x10^4N car is parked 8m from one end of the bridge how much force does each pillar exert?
...............8m
..............o--o
---------------------------------
/.../...../....../...../....../......../ |
-------------------------------
3m..|....20m........................|3m
.....F1................................F2

force down of 400,000 N
Car weighing 19,800 N(force down) is 8m from one end of the bridge.

sum force down = sum force up
400,000N+19,800N= F1+F2(the two pillars)

419800N=F1+F2

sum torque cw=sum torque ccw

torque=fd

ccw forces: bridge,car

(400,000N)(7m)+(19,800N)(5m)=F1(0)+F2(14m)

14F2=2,800,000N/M+ 99000N/M
F2=207,071.43N
Plug into the equation from the force up/force down part,

419,800N=F1+207071.43
F1=212,728.57N

7. 1.14m from the 300N child.

done.

EDIT: fatherofcajun beat me to it

8. Originally Posted by Father of Cajun
sum torque clockwise= sum torque ccw
torque=fd

so 400d1=300d2
d1+d2=2
d1=2-d2
400(2-d2)=300d2
800-400d2=300d2
800=700d2
d2=1.14m from the 300N child.

Anyone wanna check my work for me?
ur good

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