Math Problem Statement

Block A in (Figure 1) weighs 1.30 N , and block B weighs 3.60 N . The coefficient of kinetic friction between all surfaces is 0.310.

Find the magnitude of the horizontal force F⃗ necessary to drag block B to the left at constant speed if A is held at rest (figure (b)).

Block A weighs 1.30 N, and block B weighs 3.60 N. The coefficient of kinetic friction between all surfaces is 0.310. Find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed if A is held at rest (figure b).

Solution

Let's break this problem down and solve it step by step:

Given dataLet's break this problem down and solve it step by step:

Given

  • Block AA weighs 1.30N1.30 \, \text{N}.
  • Block BB weighs 3.60N3.60 \, \text{N}.
  • Coefficient of kinetic friction between all surfaces: μk=0.310\mu_k = 0.310.
  • Block AA is held at rest, and we need to find the horizontal force F\vec{F} necessary to drag block BB to the left at a constant speed.

Step 1: Calculate normal forces

The normal force exerted on each block is equal to the weight of the block because the forces are vertical.

  • For block AA, the normal force is NA=1.30NN_A = 1.30 \, \text{N}.
  • For block BB, the normal force is the sum of its own weight and the weight of block AA, since block AA is sitting on top of block BB. NB=3.60N+1.30N=4.90NN_B = 3.60 \, \text{N} + 1.30 \, \text{N} = 4.90 \, \text{N}

Step 2: Calculate friction forces

The kinetic friction force is given by: fk=μkNf_k = \mu_k \cdot N

  • Friction force between block AA and block BB: fA/B=μkNA=0.3101.30N=0.403Nf_{A/B} = \mu_k \cdot N_A = 0.310 \cdot 1.30 \, \text{N} = 0.403 \, \text{N}

  • Friction force between block BB and the surface: fB/surface=μkNB=0.3104.90N=1.519Nf_{B/\text{surface}} = \mu_k \cdot N_B = 0.310 \cdot 4.90 \, \text{N} = 1.519 \, \text{N}

Step 3: Analyze forces acting on block BB

Since block AA is held at rest, the force required to drag block BB at a constant speed is the sum of the friction forces opposing the motion:

Fnet=fA/B+fB/surface=0.403N+1.519N=1.922NF_{\text{net}} = f_{A/B} + f_{B/\text{surface}} = 0.403 \, \text{N} + 1.519 \, \text{N} = 1.922 \, \text{N}

Final Answer:

The magnitude of the horizontal force F\vec{F} necessary to drag block BB to the left at constant speed is 1.922N1.922 \, \text{N}.


Would you like a deeper explanation or further details? Here are some related questions:

  1. How would the force change if block AA were allowed to slide freely?
  2. What role does the coefficient of friction play in determining the force?
  3. How would increasing the weight of block BB affect the required force?
  4. How do we calculate normal forces in more complex systems like inclined planes?
  5. What happens to the frictional force if the surface between BB and the ground changes?

Tip: When objects are in contact, the normal force and friction play crucial roles in determining how much force is needed to move them at a constant velocity. Always account for all points of contact.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Friction

Formulas

Friction force formula: f_k = μ_k * N
Normal force formula: N = weight of the object
Net force required to drag block: F_net = f_A/B + f_B/surface

Theorems

Newton's First Law (Inertia)
Kinetic friction

Suitable Grade Level

Grades 9-12