Question

-4.44 A mass \(m_{1}=20.0 \mathrm{~kg}\) on a frictionless ramp is attached to a light string. The string passes over a frictionless pulley and is attached to a hanging mass \(m_{2}\). The ramp is at an angle of \(\theta=30.0^{\circ}\) above the horizontal. \(m_{1}\) moves up the ramp uniformly (at constant speed). Find the value of \(m_{2}\)

Short answer

Based on the step-by-step solution, create a short answer: The hanging mass m2 needs to be 10.0 kg for m1 to move up the ramp at a constant speed. We determined this by analyzing the forces acting on both masses using Newton's Second Law of Motion, resolving the gravitational force into components along and perpendicular to the ramp, and solving for m2.

Step-by-step solution

Draw a free body diagram for both masses

Begin by drawing a free body diagram to visualize the forces acting on both masses. For m1, we represent the gravitational force acting downward (mg), and the tension force (T) acting along the string. For m2, we have the gravitational force acting downwards (m2g) and the tension force (T) acting upward through the string.

Resolve the gravitational force acting on m1 into components

To resolve the gravitational force acting on m1, we need to break the force into components parallel and perpendicular to the ramp. The parallel component is \(m_{1}g\sin(\theta)\), and the perpendicular component is \(m_{1}g\cos(\theta)\), where \(\theta\) is the angle of the ramp with the horizontal.

Set up equations using Newton's Second Law of Motion

Since m1 is moving up the ramp at a constant speed, the net force acting on m1 along the ramp is zero. Therefore, we have: $$T - m_{1}g\sin(\theta) = 0$$ For m2, the net force acting on it will equal its mass times its acceleration: $$m_{2}g - T = m_{2}a$$

Solve for the acceleration

At this point, we can use the first equation to solve for the acceleration of m2: $$T = m_{1}g\sin(\theta) \Rightarrow a = \frac{T-m_{2}g}{m_{2}} = \frac{m_{1}g\sin(\theta)-m_{2}g}{m_{2}}$$ Since m1 is moving up the ramp at a constant speed, the acceleration a must be equal to zero. Therefore: $$\frac{m_{1}g\sin(\theta)-m_{2}g}{m_{2}} = 0$$

Solve for m2

Now we can solve the above equation for m2: $$m_{1}g\sin(\theta)-m_{2}g = 0 \Rightarrow m_{2} = \frac{m_{1}g\sin(\theta)}{g}$$ Plugging in the given values for m1 and theta: $$m_{2} = \frac{20.0\,\text{kg}\cdot 9.81\,\text{m/s}^2\sin(30.0^\circ)}{9.81\,\text{m/s}^2}$$ Calculating the value of m2: $$m_{2} = 10.0\,\text{kg}$$ So, the hanging mass m2 needs to be \(10.0\,\text{kg}\) for m1 to move up the ramp at a constant speed.

Key concepts

Understanding Free Body Diagrams

A free body diagram is a powerful tool in physics to visualize all the forces acting upon a single object. It's like drawing a map of all the force-vectors that comes into play for that object. In our textbook exercise, we draw separate diagrams for two masses: one on a ramp and one hanging.

In the diagram for the mass on the ramp, we see two forces: the weight of the mass acting directly downwards, represented by the force vector labeled as mg, and the tension in the string, which we'll call 'T'. This tension is what's keeping the mass from sliding down the ramp due to gravity. For the hanging mass, the forces are its weight (labeled as m2g), pulling down, and the tension 'T' pulling up. Creating a free body diagram is typically the first and a crucial step to solving problems in mechanics because it helps us visualize and thus better understand the forces at play.

The Dynamics of a Frictionless Ramp and Pulley System

A frictionless ramp and pulley system simplifies the complex real-world interactions into a more manageable problem by ignoring the effects of friction. This assumption allows us to focus solely on the gravitational and tension forces. When we say the ramp is frictionless, we're imagining a very smooth surface where the only force resisting the mass's movement up or down the ramp is its own weight component along the ramp, without any additional frictional forces.

Similarly, a frictionless pulley means that there is no resistance in the pulley's rotation, which in the real world could be caused by the axle or wheel rubbing against other surfaces. This ideal situation ensures that the tension in the rope is the same on either side of the pulley – a fact that simplifies calculations and lets students focus on the fundamental principles of mechanics without getting bogged down by complex friction calculations.

Components of Gravitational Force on an Inclined Plane

In physics problems, we often break down forces into components to simplify calculations. The component of gravitational force is a prime example. On an inclined plane, like our ramp, gravity pulls the mass both down towards the center of the Earth and along the plane.

Mathematically, we split the gravitational force into two parts: the component perpendicular to the ramp, which does not contribute to movement and is represented by \(m_{1}g\cos(\theta)\), and the parallel component, \(m_{1}g\sin(\theta)\), that causes the mass to slide down the ramp or requires tension in the string to resist this sliding. Understanding this concept is key to solving physics problems involving inclined planes, as it allows us to focus on the forces that actually cause movement.