Question

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.35 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.

Answer 1

Answer:

$T_{1}$ = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about $T_{1}$ we have:

(M + m) g * 0.5L - $T_{2}$(L - d) = 0

⇒ $T_{2}$ = [(M + m) g * 0.5L] ÷ (L - d)

$T_{2}$ = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

$T_{2}$= 59.535 ÷ 2.4

$T_{2}$ = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

$T_{1}$ + $T_{2}$ = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

$T_{1}$ + 24.81 = 26.46 + 13.23

$T_{1}$ = 14.88 N