Question

A 1000.-N crate of length \(L\) rests on a horizontal platform. It is being pulled up by two vertical ropes. The left rope has a tension of \(400 . \mathrm{N}\) and is attached a distance of \(L / 4\) from the left end of the crate. Assuming the platform moves downward and the crate does not move or rotate, what can you say about the tension in the right rope?

Short answer

Answer: The tension in the right rope is 200 N.

Step-by-step solution

Analyze the forces acting on the crate

There are three forces acting on the crate: the force acting downwards due to gravity (weight W=1000 N), the tension in the left rope (T1=400 N), and the tension in the right rope (T2, which we need to find). The weight of the crate acts at its center of mass, i.e., at L/2 from the left end. The left rope exerts force a distance L/4 from the left end, and the right rope exerts force at the right end of the crate.

Write the equations for static equilibrium

For the crate to be in static equilibrium, both the sum of the forces and the sum of the torques acting on the crate must be equal to zero. Let's first write the equation for the net force: F_net = T1 + T2 - W = 0. For the torques, we will choose the left end as the reference point (denoted as A). The equation for the net torque is τ_net = (T1 * 0) - (W * L/2) + (T2 * L) = 0. Note that the torque due to the weight is negative because it acts in the clockwise direction, and the torque due to T2 is positive because it acts in the counter-clockwise direction.

Solve the equations for the tension in the right rope

From the equation of the net force, we can write T2 = W - T1. Now, substitute this expression into the torque equation: τ_net = - (W * L/2) + ((W - T1) * L) = 0. Simplify the equation to get: L = (T1 * L) / (W/2). Now, solve for the tension T2 = W - T1 = W - (W * L / 2). Plug in the values of W and L to find the tension in the right rope: T2 = 1000 - ((1000 * L / 4) / (1000/2)) = 1000 - 2000 / (L/4) = 1000 - 800 = 200 N. Therefore, the tension in the right rope is 200 N.

Key concepts

Tension in Ropes

Understanding the tension in ropes is crucial when analyzing systems in static equilibrium. Tension is a pulling force that is transmitted through a flexible connector, such as a rope or cable, when it is pulled tight by forces acting at each end.
In our example with the 1000-N crate, two ropes work to balance the weight of the crate in the static system. Since the left rope has a tension of 400 N, we use the principle that the sum of the forces must be zero for equilibrium to determine the tension in the right rope. This directly reflects Newton’s Third Law of Motion - for every action, there is an equal and opposite reaction.
In simpler terms, the rope tensions counterbalance the weight of the crate, preventing it from accelerating. Hence, the tension force in the right rope, combined with the tension force in the left rope, must equal the weight of the crate to maintain static equilibrium.

Torque and Rotational Equilibrium

Let's delve into the concept of torque and rotational equilibrium. Torque, often symbolized as \( \tau \), is a measure of the force that can cause an object to rotate about an axis. It is the product of force and the distance from the point of rotation, known as the moment arm.
In the case of the crate, each tension force generates a torque about a point, which in this exercise, is selected to be the left end of the crate. When an object is in rotational equilibrium, the net torque acting on it is zero, meaning there is no rotational acceleration and the object remains stationary or rotates at constant velocity.

Calculating Torque

For torque calculations, we utilize the equation \( \tau = r \times F \) where \( r \) is the moment arm and \( F \) is the force. In this problem, the torque generated by the weight acts to rotate the crate clockwise, and the torque from the right rope acts to rotate it counter-clockwise. By setting the net torque to zero, we can solve for the unknown tension in the right rope using the distances provided from the pivot point.

Forces in Physics

In physics, 'force' is any interaction that, when unopposed, will change the motion of an object. Forces can be categorized in different ways: applied force, gravitational force, tension force, and many others.
In our crate scenario, the gravitational force (weight \( W \)) acts vertically downward, and it is opposed by two vertical tension forces (\( T1 \), \( T2 \) - the forces in the ropes). The weight is constant at 1000 N since it is essentially equal to the gravitational pull on the crate. Ensuring the sum of these forces equals zero guarantees the crate's static equilibrium in the vertical direction, preventing any vertical motion.
These relationships are fundamental to Newton’s laws of motion, which describe the relationship between a body and the forces acting upon it, and the body's motion in response to those forces.

Net Force and Net Torque Calculations

Net force and net torque calculations are the backbone of solving static equilibrium problems. The net force is the vector sum of all the individual forces acting on a body. For a system to be in static equilibrium, the net force must be zero.
Similarly, the net torque is the sum of all individual torques about a given axis. To achieve rotational equilibrium, the net torque must also be zero. This condition ensures no rotational motion occurs.

Solving for Unknowns

In this exercise, after establishing that the net force on the crate is zero (\( F_{net} = T1 + T2 - W = 0 \) ), we calculate the net torque to be zero at the pivot point (\( \tau_{net} = (T1 \cdot 0) - (W \cdot L/2) + (T2 \cdot L) = 0 \) ). We then combine these findings to solve for the unknown tension \( T2 \) in the right rope. This process allows us to conclude that the tension in the right rope must be 200 N to maintain the crate in static equilibrium. Such techniques are widely applicable in engineering, constructions, and various realms of physics.