To find the magnitude of the components of the block's weight perpendicular and parallel to the incline, we will apply some basic trigonometry.
A block resting on an incline will have two components of force due to gravity:
1. The perpendicular component, which is the force exerted by the block perpendicular to the incline.
2. The parallel component, which is the force causing the block to slide down the incline.
Let's denote:
- \( w \) as the weight of the block, which is 90 lb in this case, and
- \( \theta \) as the angle of incline, which is 15°.
**Finding the perpendicular component**:
The perpendicular component (\( w_{\perp} \)) is given by the formula:
\[ w_{\perp} = w \cdot \cos(\theta) \]
In this case:
\[ w_{\perp} = 90 \cdot \cos(15°) \]
After performing the calculation (using a calculator with the cosine function), the perpendicular component of the weight is approximately 86.9 lb.
**Finding the parallel component**:
The parallel component (\( w_{\parallel} \)) is obtained using the sine function as follows:
\[ w_{\parallel} = w \cdot \sin(\theta) \]
Thus:
\[ w_{\parallel} = 90 \cdot \sin(15°) \]
Using a calculator to find the sine of 15° and multiply it by the weight of the block, we find that the parallel component of the weight is approximately 23.3 lb.
In conclusion:
- The magnitude perpendicular to the incline is 86.9 lb.
- The magnitude parallel to the incline is 23.3 lb.
Both values are rounded to the nearest tenth as required.
