physics tension

Question

Two blocks are connected by a massless string and are held in position by another massless string along a frictionless incline (as shown in the figure). Let M₁ = 2.30 kg, M₂ = 4.70 kg, and θ� = 20.0°. Calculate the tension in the string connecting the two blocks.

What is the tension in the string which connects mass M₂ to the wall?

Answer 1

For this problem you need to draw free body diagrams and use them to write equations using Newton's second law.

a) The tensions of the string between the blocks.

Use coordinates with the x-axis going up the ramp and the y-axis perpendicular to the ramp. You will need to break the force of gravity into components in the new coordinate system.

F_g = -sin(θ) m₁g i + cos(θ) m₁g j

As Newton's second law is a vector equation each component of the of the vectors make their own equation. Consider the free body diagram on m₁ and let T₁₂ be the tension in the string between the blocks.

x-direction

T₁₂ - m₁g sin(θ) = m₁ a_x = 0

T₁₂ = m₁g sin(θ) = (2.30 kg)(9.8 m/s²) sin(20º) ≈ 7.71 N

b) Find the tension in the second cord.

Using the free body diagram on m₂

x-direction

-T₁₂ - m₂g sin(θ) + T = ma_x = 0

T = T₁₂ + m₂g sin(θ) = (m₁ + m₂)g sin(θ) ≈ 26.81 N