To solve for the tension $T$ required to keep the block at rest on the inclined plane, we'll break down the forces acting on the block.

Step 1: Identify the Forces

The block has a weight $W$ of 31 lb acting vertically downward. This weight can be resolved into two components:

1. Parallel component to the inclined plane: $W_{\parallel}$
2. Perpendicular component to the inclined plane: $W_{\perp}$

Step 2: Resolve the Weight into Components

1. Parallel Component $W_{\parallel}$: $W_{\parallel} = W \times \sin(\theta) = 31 \times \sin(18^\circ)$

2. Perpendicular Component $W_{\perp}$: $W_{\perp} = W \times \cos(\theta) = 31 \times \cos(18^\circ)$

Since the plane is friction-free, the perpendicular component does not affect the tension in the rope.

Step 3: Equilibrium Condition

To keep the block at rest, the tension $T$ in the rope must exactly balance the parallel component of the weight $W_{\parallel}$.

So, the tension $T$ required is: $T = W_{\parallel} = 31 \times \sin(18^\circ)$

Step 4: Exact Value of Tension

The exact value of $T$ can be left in terms of $\sin(18^\circ)$, or you can calculate the numerical value.

$T = 31 \times \sin(18^\circ)$ $\sin(18^\circ) \approx 0.3090$ $T \approx 31 \times 0.3090 = 9.579 \, \text{lb}$

Final Answer

The exact tension $T$ required to keep the block at rest is: $T = 31 \times \sin(18^\circ) \, \text{lb}$ The approximate value is: $T \approx 9.579 \, \text{lb}$

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Would you like more details or have any other questions?

Related Questions

1. How would the tension change if the angle of the inclined plane increased?
2. What would happen to the tension if the plane had friction?
3. How do we find the normal force acting on the block in this scenario?
4. What if the rope were attached at a different angle?
5. How does the mass of the block affect the tension in the rope?

Tip:

When resolving forces on an inclined plane, always check that you are using the correct trigonometric functions (sine for the parallel component and cosine for the perpendicular component).