0.14 Phy1110: force -- moments, torque, couple, and equilibrium  (Page 10/13)

Simplifying, rearranging terms, and dividing both sides by 2 gives us:

P = 4*Q ( eq. a2 )

Two equations and two unknowns

Including eq. a1 , (repeated below) we now have two equations and two unknowns.

P + Q = W

Inserting the numeric value for W gives us

P + Q = 10, or

Q = 10 - P ( eq. a3 )

Substituting this value of Q into eq. a2 gives us

P = 4*(10 - P), or

P = 40 - 4*P, or

5*P = 40, or

P = 8

which is one of the answers that we are looking for.

Inserting the value for P into eq. a3 gives us

Q = 10 - 8, or

Q = 2

which is the other answer that we are looking for.

A more general case of the trapeze bar

Let's pick another point, label it X, and compute the moments about that point. Those moments must also sum to zero for the bar to be in equilibrium.(The moments computed about any point on the bar must sum to zero for the bar to be in equilibrium.)

The moments about X produced by the three forces are:

• P: (X-A)*(P)
• W: (X-C)*(W)
• Q: (X-B)*(Q)

Let X = 5

Substituting for X gives us:

(5)*(P) - (3)*(10) + (-5)*(Q) = 0

Simplifying, rearranging terms, and diving both sides by 5 gives us:

P - Q - 6 = 0, or

P = Q + 6 ( eq. a4 )

Now we can use this equation along with eq.a1 to solve for Q.

P + Q = 10, from eq.a1 , or

Q = 10 - P

Substituting this value for Q in eq. a4 gives us,

P = 10 - P + 6

Simplifying and dividing both sides by 2 gives us,

P = 8 , which is the same answer as before (which it should be)

Substitution of P back into eq. a1 gives us,

Q = 2

Apply theweight at different locations on the trapeze

If you solve for P and Q for any location of the weight between the ropes, you will find that the values for the upward forces at each end of the bar areinversely proportional to the distance from the weight to that end of the bar.

For example, for equilibrium, using the dimension symbols established for your drawing earlier:

a*P = b*Q

Dividing both sides by a gives:

P = (b/a)*Q

Weight is centered

If b/a = 1 (weight is centered), then:

P = Q meaning that both upward forces are equal.

Weight towards the left end

If b/a = 4 (weight at 2), then:

P = 4*Q meaning that most of the force is being exerted by the rope on the left end.

Weight toward the right

If b/a = 1/4 (weight at 8), then

P = Q/4 meaning that most of the force is being exerted by the rope on the right end.

Weight at the right end

If b/a = 0 (weight at 10, right end of the bar)

P = 0 meaning that all of the force is being exerted by the rope on the right end of the bar.

Because P + Q = W, we conclude that Q = W

The bar is essentially eliminated

The scenario where b/a = 0 essentially eliminates the bar from consideration. The weight is hanging directly on the rope on the right end of the bar and theweightless bar and the other rope are simply floating in the air.

Hypothetical replacement by a single upward force

If we were to draw an imaginary force labeled R pointing directly up from the point C where R is equal to P + Q, we could imagine the forces P and Q as beingreplaced by R. In that case, R would be the resultant of P and Q and would be equal in magnitude and opposite in direction to the downward force W.