Math Problem Statement

Suppose the angles shown in the figure are 54 ∘ and 25 ∘. If the left-hand mass is 1.7 kg ,what should the right-hand mass be so that it accelerates downslope at 0.64 m/s2 ? Express your answer with the appropriate units.

Suppose the angles shown in the figure are 54° and 25°. If the left-hand mass is 1.7 kg, what should the right-hand mass be so that it accelerates downslope at 0.64 m/s²? Express your answer with the appropriate units.

Solution

To solve this problem, we'll break it down step by step, using Newton's laws and the given details to determine the unknown mass.

We are given the following information:

  • Left-hand mass (m1m_1) = 1.7 kg
  • Right-hand mass (m2m_2) = ?
  • Acceleration (aa) = 0.64 m/s²
  • Angles: 5454^\circ for the left-hand incline, and 2525^\circ for the right-hand incline

Step 1: Write down the force components along the slopes for both masses

For each mass, the forces acting along the incline include gravity, tension in the string, and friction (if applicable, but none is mentioned here).

Left-hand mass (m1m_1) on the 5454^\circ slope:

  • Force due to gravity: Fg=m1gsin(54)F_g = m_1 g \sin(54^\circ)
  • Tension in the string: TT
  • Net force: Fnet=m1a=m1gsin(54)TF_{net} = m_1 a = m_1 g \sin(54^\circ) - T

Right-hand mass (m2m_2) on the 2525^\circ slope:

  • Force due to gravity: Fg=m2gsin(25)F_g = m_2 g \sin(25^\circ)
  • Tension in the string: TT
  • Net force: Fnet=m2a=Tm2gsin(25)F_{net} = m_2 a = T - m_2 g \sin(25^\circ)

Step 2: Set up the system of equations

Since the tension is the same for both masses, we can set the two net forces equal but opposite in sign:

  1. m1gsin(54)T=m1am_1 g \sin(54^\circ) - T = m_1 a
  2. Tm2gsin(25)=m2aT - m_2 g \sin(25^\circ) = m_2 a

Step 3: Solve for the tension TT

From equation (1): T=m1gsin(54)m1aT = m_1 g \sin(54^\circ) - m_1 a

Step 4: Substitute TT into equation (2)

Substitute the expression for TT into equation (2): (m1gsin(54)m1a)m2gsin(25)=m2a(m_1 g \sin(54^\circ) - m_1 a) - m_2 g \sin(25^\circ) = m_2 a

Step 5: Solve for m2m_2

Now, we can solve for m2m_2: m1gsin(54)m1a=m2gsin(25)+m2am_1 g \sin(54^\circ) - m_1 a = m_2 g \sin(25^\circ) + m_2 a

Let me calculate the value of m2m_2.The right-hand mass (m2m_2) should be approximately 2.59kg2.59 \, \text{kg} in order for the system to accelerate downslope at 0.64m/s20.64 \, \text{m/s}^2.

Would you like further details or clarification on any of the steps?

Here are some related questions you might find useful:

  1. What happens if the angles of the inclines change?
  2. How does the system behave if there is friction on the incline?
  3. Can the acceleration be negative, and how would that affect the system?
  4. How does changing one of the masses affect the acceleration?
  5. What if the pulley has mass or friction? How would the equations change?

Tip: Always double-check the angle conversion when solving problems involving slopes and inclines in trigonometric functions!

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
Inclined Plane
Trigonometry

Formulas

Net Force: F_net = m * a
Force due to gravity on incline: F_g = m * g * sin(θ)
Tension in the string and system of equations

Theorems

Newton's Second Law

Suitable Grade Level

Grades 11-12