Question

3 - 38. the block has a weight of 20 lb and is being hoisted at uniform velocity. determine the angle θ for equilibrium and the force in each cord. 3 - 39. determine the maximum weight w of the block that can be hoisted at constant velocity if each cord can support a maximum tension of 80 lb. also, what is the angle θ for equilibrium? problems 3 - 38/39

Explanation:

Step1: Analyze forces at pulley A

Since the block is hoisted at uniform velocity, the system is in equilibrium. The tension in the cord is the same throughout the cord due to the nature of an in - extensible cord passing over a frictionless pulley. Let the tension in the cord be $T$.

Step2: Write force equilibrium equations for problem 3 - 38

For the equilibrium of the pulley A, $\sum F_x=0$ and $\sum F_y = 0$. The weight of the block $W = 20$ lb.
In the x - direction: $T\cos\theta=T\cos30^{\circ}$, which implies $\theta = 30^{\circ}$.
In the y - direction: $T\sin\theta+T\sin30^{\circ}=W$. Substituting $\theta = 30^{\circ}$ and $W = 20$ lb, we have $T\sin30^{\circ}+T\sin30^{\circ}=20$. So $2T\times\frac{1}{2}=20$, and $T = 20$ lb.

Step3: Write force equilibrium equations for problem 3 - 39

The maximum tension in each cord is $T = 80$ lb.
In the y - direction: $T\sin\theta+T\sin30^{\circ}=W$. In the x - direction: $T\cos\theta=T\cos30^{\circ}$, so $\theta = 30^{\circ}$.
Substituting $T = 80$ lb and $\theta = 30^{\circ}$ into the y - direction equation: $80\times\frac{1}{2}+80\times\frac{1}{2}=W$. So $W=80$ lb.

Answer:

For problem 3 - 38: $\theta = 30^{\circ}$, $T = 20$ lb
For problem 3 - 39: $W = 80$ lb, $\theta = 30^{\circ}$