Math Problem Statement

A block is pushed up a plane with a constant force of 50 N parallel to the plane. The plane is inclined at 30° to the horizontal, and the block has a weight of 100 N. The coefficient of friction between the block and the plane is 0.25. The task is to find the magnitude of the normal force exerted by the plane on the block.

Solution

The image contains a physics problem where a block on an inclined plane is subjected to various forces. Here's a breakdown of the text and diagram:

Problem Statement: A block is pushed up a plane with a constant force of 50 N parallel to the plane. The plane is inclined at 30° to the horizontal, and the block has a weight of 100 N. The coefficient of friction between the block and the plane is 0.25. The task is to find the magnitude of the normal force exerted by the plane on the block.

Step-by-step solution:

  1. Identify forces:

    • Weight of the block: W=100NW = 100 \, \text{N}
    • Applied force parallel to the plane: Fapplied=50NF_{\text{applied}} = 50 \, \text{N}
    • Incline angle: θ=30\theta = 30^\circ
    • Coefficient of friction: μ=0.25\mu = 0.25
  2. Decompose the weight:

    • The weight can be decomposed into two components:
      • Perpendicular to the plane (Normal force): W=Wcos(θ)=100cos(30)W_\perp = W \cdot \cos(\theta) = 100 \cdot \cos(30^\circ)
      • Parallel to the plane (along the incline): W=Wsin(θ)=100sin(30)W_\parallel = W \cdot \sin(\theta) = 100 \cdot \sin(30^\circ)
  3. Normal force calculation: The normal force NN is the force perpendicular to the plane. It must balance the component of the weight perpendicular to the plane and any other vertical forces. Since the applied force is parallel to the plane, it does not contribute to the normal force. Thus: N=W=Wcos(30)N = W_\perp = W \cdot \cos(30^\circ) N=100cos(30)N = 100 \cdot \cos(30^\circ) Using cos(30)=320.866\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866, we get: N1000.866=86.6NN \approx 100 \cdot 0.866 = 86.6 \, \text{N}

Thus, the magnitude of the normal force is approximately 86.6 N.

Would you like any further details, or do you have any specific questions? Here are five questions to deepen your understanding:

  1. How is the weight of the block decomposed into parallel and perpendicular components on an inclined plane?
  2. How does the angle of the incline affect the normal force acting on the block?
  3. What role does friction play in determining the net force on the block along the plane?
  4. How would the normal force change if the incline angle were increased or decreased?
  5. How is the coefficient of friction used to calculate the force of friction in similar problems?

Tip: When working with inclined planes, always start by resolving forces into components parallel and perpendicular to the surface.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Forces on Inclined Planes
Normal Force
Decomposition of Forces

Formulas

Weight component perpendicular to the plane: W_perp = W * cos(θ)
Weight component parallel to the plane: W_parallel = W * sin(θ)
Normal Force: N = W * cos(30°)

Theorems

Newton's Second Law of Motion
Friction Force

Suitable Grade Level

Grades 9-12