Figure 11-16a shows a safe (mass M = 430 kg) hanging by a rope (negligible mass) from a boom (a = 1.9 m and b = 2.5 m) that consists of a uniform hinged beam (m=85 kg) and horizontal cable (negligible mass).
Find the magnitude F of the net force on the beam from the hinge.

Explanation

Now we want the horizontal component `F_(h)` and vertical component F, so that we can combine them to get the magnitude `F_(v)` of the net force.Because we know `T_(c)`, we apply the force balancing equations to the beam.
Calculations: For the horizontal balance, we can rewrite `F_("net,x")=0`
`F_(h)-T_(c)=0`
and so `F_(h)= T_(c)= 6093 N`
For the vertical balance, we write `F_("net,x")=0` as
`F_(v)-mg-T_(r)=0`
Substituting Mg for `T_(r)` and solving for `F_(v)`, we find that
`F_(v)= (m+M)g= (85kg+430kg)(9.8m//s^(2))`
= 5047N
From the Pythagorean theorem, we now have
`F= sqrt(F_(h)^(2)+ F_(v)^(2))`
`= sqrt((6093)^(2)+ (5047N)^(2)) ~~ 7900N`
Note that F is substantially greater than either the combined weights of the safe and the beam, 5000 N, or the tension in the horizontal wire, 6100 N.