The block has a weight of 30 lb and is being hoisted at uniform velocity. Suppose that γ = 25 ∘. Determine the angle θ for equilibrium.
see figure here: session.masteringengineering.com/problemAsset/2086482/4/Hibbeler14.ch3.p10.jpg
To determine the angle θ for equilibrium of the block being hoisted, we can use the principles of static equilibrium. In this case, we need to consider the forces acting on the block.
From the given figure, we can see that the weight of the block (W) acts vertically downward. We can resolve this weight force into its components along the x-axis (Wx) and the y-axis (Wy) using trigonometry.
Wx = W * cos(γ)
Wy = W * sin(γ)
Since the block is being hoisted at uniform velocity, the net force in both the x and y direction should be zero. This means that the sum of the vertical forces and the sum of the horizontal forces should be equal to zero.
Sum of vertical forces = Wy
Sum of horizontal forces = T - Wx
Since the block is in equilibrium, the sum of vertical forces and the sum of horizontal forces must equal zero.
Wy + (T - Wx) = 0
We can substitute the values of Wy and Wx:
W * sin(γ) + (T - W * cos(γ)) = 0
Now, we can solve for the tension force (T) in terms of the known values (W, γ):
T = W * cos(γ) - W * sin(γ)
Finally, to find the angle θ for equilibrium, we need to determine the angle that satisfies this tension force equation. This can be done by solving the equation for T = 0 and finding the angle at which T is zero.
θ = arctan(W * sin(γ) / (W * cos(γ)))
Now we can substitute the given values (W = 30 lb, γ = 25°) into the equation to find the angle θ for equilibrium.