To find the ratio of forces around a right triangle

[gnits]

Homework Statement

To find the ratio of forces around a right triangle

Relevant Equations

Equate moments of couples

Could I please ask for help with the following:

ABC is a right-angled triangle in which AB = 4a; BC = 3a. Forces of magnitudes P, Q and R act along the directed sides AB, BC and CA respectively. Find the ratios P:Q:R if their resultant is a couple.

Book answer is 4 : 3 : 5

Here's my diagram:

Here's my working:

tan ACB = 4/3 and so the cos of ACB is 3/5 and the sin of ACB is 4/5

As we are given that the resultant is a couple we can equate moments about A, B and C respectively, giving:

4aQ = 4R/5 * 3a = 3aP

Which gives:

20Q = 12R = 15P

so P : Q : R = 15 : 20 : 12

Not the book answer,
thanks for pointing to my error,

Mitch.

 

[vela]

20Q = 12R = 15P

so P : Q : R = 15 : 20 : 12

Say P=15. Then Q=20. But 20Q = 400 and 15P=225, so $20Q \ne 15P$.

 

[gnits]

Thanks, I see my error.

Mitch.