Question

eg2. a crate with a mass of 12.5 kg sits on a frictionless surface at an angle of 20° to the horizontal. it is connected to second crate with a mass of 10.0 kg by a string that passes over a pulley, as shown in the system diagram.
a) calculate the magnitude of the acceleration of the system of crates.
b) determine the magnitude of the tension in the string.

Explanation:

Step1: Analyze forces and apply Newton's second law

Let \(m_1 = 12.5\space kg\) (crate on the incline) and \(m_2=10.0\space kg\) (hanging crate).
For the crate on the incline (\(m_1\)): \(F_{net1}=m_1g\sin\theta - T=m_1a\)
For the hanging crate (\(m_2\)): \(F_{net2}=T - m_2g=m_2(-a)\) (negative because it moves upward if \(m_1\) moves down the incline)
Add the two equations: \(m_1g\sin\theta - m_2g=(m_1 + m_2)a\)

Step2: Calculate acceleration \(a\)

Given \(g = 9.8\space m/s^2\), \(\theta = 20^{\circ}\), \(m_1 = 12.5\space kg\), \(m_2 = 10.0\space kg\)
\(a=\frac{m_1g\sin\theta - m_2g}{m_1 + m_2}\)
\(a=\frac{12.5\times9.8\times\sin20^{\circ}- 10.0\times9.8}{12.5 + 10.0}\)
\(\sin20^{\circ}\approx0.342\)
\(a=\frac{(12.5\times9.8\times0.342)-98}{22.5}\)
\(a=\frac{(41.415)-98}{22.5}=\frac{- 56.585}{22.5}\approx - 2.51\space m/s^2\) (magnitude \(|a|\approx2.5\space m/s^2\))

Step3: Calculate tension \(T\)

From \(T - m_2g=m_2(-a)\)
\(T=m_2(g - a)\)
Substitute \(m_2 = 10.0\space kg\), \(g = 9.8\space m/s^2\), \(a\approx2.5\space m/s^2\)
\(T = 10\times(9.8 - 2.5)=10\times7.3 = 73\space N\)

Answer:

a) The magnitude of the acceleration of the system of crates is approximately \(2.5\space m/s^2\)
b) The magnitude of the tension in the string is \(73\space N\)