# Physics hw. our teacher offered us 5 extra pts on our test on mon. if we get c right

1.AB is a uniform bar of length 5m and weight 60N. It is supprted in a horizontal position by 2 vertical strings P and Q. P is 0.5m from A
and Q 1.5m from B.

a) What are the tensions in the strings?
b) What weight must be applied at A if the tensions in the strings are to be equal?
c) What is the biggest mass which can be hung from B if the bar is to remain horizontal?

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this h.w is for 5 extra points on a test on monday if i get them right...please help.

1.AB is a uniform bar of length 5m and weight 60N. It is supprted in a horizontal position by 2 vertical strings P and Q. P is 0.5m from A
and Q 1.5m from B.

a) What are the tensions in the strings?
b) What weight must be applied at A if the tensions in the strings are to be equal?
c) What is the biggest mass which can be hung from B if the bar is to remain horizontal?

a) M(P) :

3*q -2*60 = 0
3q-120=0
q=40

[resolving vertically upwards]:
p+q-60=0
p+40-6=0
p-20=0
p=20
********************************************************************
b) finding P:

60-4.5p+5F=0
(-40 + 2p-2.5F=0)*2

60-4.5p + 5F= 0
-80 + 4p-5F=0

-20 + .5p = 0
.5p = 20

p = 20/.5
p= 40N

finding F:

40-60+40-F = 0
20-F=0
F=20N
******************************************************************

M(F):
B-40+60-40+20=0
B= 0

Last edited:

You draw all the forces and in each case you write down the following two equations:
Total force up=Weight
Total torque counterclockwise about center of mass=0
(you have to pay attention to the *signs* since the forces are taken "up" as positive and the torque "counterclockwise" about the center as positive [you can also chose clockwise if you like, but at least all torques in the *same* direction about the center of mass])
The torque is (distance from center of mass) times (perpendicular force in counterclockwise direction). For example if the force on string P is called P and string P is 2m from the center of mass, then the torque is -2*P (negative since its not counterclockwise as we agreed). Likewise Q is 1m from the center so the torque is 1*Q.

a)
forces and torque equations:
P+Q=60
-2*P+1*Q=0
Solve this set of equations and get P=20N and Q=40N

b)
again forces and torque this time with a new force A at point A (and distance 2.5 from center):
P+Q-A=60
2.5*A-2*P+1*Q=0
and P and Q should be equal so
P=Q
Solving these equations (use P=Q first to get rid of Q and continue) you get
A=15N and P=Q=37.5N

c)
this time the extra force is B at B:
P+Q-B=60
-2*P+1*Q-2.5*B=0
for the bar to stay horizontal you want string P to pull and not push (which a string cant do). so
P>0
canceling Q in the first two equations you get an equation for P and pluging this in into the last you get
20-0.5*B>0
This corresponds to
B<40N
which is the maximum weight you can apply at B.

Thanks so much :D:D:D