A 525-N trunk is placed on an inclined plane that forms an angle of 30.0 degrees with the horizontal. What are the components of the weight parallel to the plane and normal to the plane?
Memorize this:
weight normal = mg cosTheta
weight parallel= mg Sin Theta.
Normal = 454.7 lbs
Parallel = 263 lbs
Well, if your equations are correct (which they are), then simple multiply 525 N, which is mg or weight, by the cosine of 30, then by the sine of 30.
weight normal is about 81 and weight parallel is about -519
To find the components of the weight parallel and normal to the inclined plane, you can use the formulas mentioned. Here's how:
Step 1: Determine the mass of the trunk. The trunk's weight is given as 525 N, which is the force due to gravity acting on the trunk. To find the mass (m), use the formula: weight = mass × gravitational acceleration (g). In this case, the value of g is approximately 9.8 m/s^2.
525 N = m × 9.8 m/s^2
Dividing both sides of the equation by 9.8 m/s^2 will give you the mass:
m = 525 N / 9.8 m/s^2
Step 2: Calculate the weight component normal to the plane. Use the formula weight normal = mass × g × cos(theta), where theta (θ) is the angle the inclined plane makes with the horizontal.
weight normal = m × g × cos(theta)
Substituting the known values:
weight normal = (mass from Step 1) × 9.8 m/s^2 × cos(30.0 degrees)
Step 3: Calculate the weight component parallel to the plane. Use the formula weight parallel = mass × g × sin(theta).
weight parallel = m × g × sin(theta)
Substituting the known values:
weight parallel = (mass from Step 1) × 9.8 m/s^2 × sin(30.0 degrees)
By plugging in the values and evaluating the trigonometric functions, you can find the components of the weight parallel and normal to the inclined plane.