Blocks of mass m₁ and m₂ are connected b...

Question

Blocks of mass m₁ and m₂ are connected by a massless string that passes over the pulley in FIGURE P12.64. The pulley turns on frictionless bearings. Mass m₁ slides on a horizontal, frictionless surface. Mass m₂ is released while the blocks are at rest. Suppose the pulley has mass m_p and radius R. Find the acceleration of m₁ and the tensions in the upper and lower portions of the string. Verify that your answers agree with part a if you set m_p = 0.

Accepted Answer

Hello, let's go through this practice problem. The figure shows an iron slab of mass 4 kg attached to a box having a mass of 5 kg via a massless string. The pulley connecting them has a mass of 100 50 g and a radius of 10 centimeters. Calculate the magnitude of acceleration experienced by the iron slab if a force pulled of 25 new ones is applied to it toward the left. Also calculate the magnitude of tensions in different parts of the string. Consider the pulley and the horizontal surface to be frictionless. Option A acceleration equals 2.6 m per second squared. T sub one equals 36 new ones and T sub two equals 36 new ones. Option B 2.6 m per second squared, 36 new ones and 13 new ones. Option C 2.6 m per second squared, 13 new ones and 36 new ones. D 2.8 m per second squared, 14 new ones and 14 new ones. E 2.8 m per second squared, 14 new ones and 36 new ones and option F 2.8 m per second squared. 36 new ones and 14 new ones. Let's start by defining some variables and drawing some force vectors on our diagram. So ill define the part of the rope connected to the slab to be one and will define the part of the rope connected to the box to be two. So acting on the slab to the right is the force from the tension of the rope. So we'll call that T sub one and then acting upwards on the box is going to be an upwards tension force from that section of the rope. So T sub two tea sub two, also acting on the box is the downward force due to its weight. So F sub G for box for part two or alternatively, the mass of the box M sub two multiplied by the acceleration due to gravity, which is the formula for weight. It also should be noted that we can't just ignore the pulley because the pulley has a mass that's given to us as well. So let's consider the forces acting on the pulley which are going to come from the tension of the rope. So to the left, there's going to be T sub one, this is equal to the T sub one that we discussed before due to N Newton's third law. And there's also going to be a downwards force from the rope from T sub two. However, because these two tension forces are acting along the edges of the pulley at the radius. These will actually be more of a torque rather than a force due to the radius of the pulley, which is also given in the problem. Now, before we go any further, an important constraint to be aware of is that as long as we assume that the string remains taught, meaning there's no bending, then the slab and the box are accelerating together. So this means that the acceleration of the slab and the acceleration of the box should always be equal to one another. So I'm just going to use the variable A to refer to the acceleration of both the slab and the box. Now let's apply Newton's second law to write out some force equations. Let's start with the slab, the force on the slab, we can see there is going to be a 25 Newton force acting to the left. So 25 new ones and then opposing that force is going to be t of one. So I'll use a minus sign because they're negating one another. And those are the only two forces acting in the horizontal direction on the slab. And according to Newton's second law, the net sum of all the forces acting on an object is equal to its mass multiplied by its acceleration. So we can turn this into a formula for the tension using some pretty simple algebra telling us that T sub one is equal to 25 new ones minus M sub one. A let's also apply Newton's second law to the pulley as it applies to torque. So recall that torque is equal to the force acting on the object multiplied by the radius or the radial distance from the point where the force is acting to the point of rotation, which is just the radius of the pulley itself. So one of the torques acting on the pulley is going to be from T sub one multiplied by the Poly's radius. But opposing that is also the torque from T sub two. The tub two again multiplied by the same radius. And according to Newton's second law for rotation, this is going to be equal to the moment of inertia for the pulley multiplied by the angular acceleration of the pulley. We can turn this equation into a relationship between the tensions. If we factor out the R so R multiplied by T sub one minus T sub two and then divide both sides of the equation by R. So that the equation can become T sub one minus T sub two equals I alpha divided by the radius. Also recall that angular acceleration is equal to linear acceleration divided by radius. So if we substitute that in for our 10 formula, then we can rewrite this as the moment of inertia multiplied by the regular acceleration divided by R squared, giving us a relationship between the two tensions and the linear acceleration. Since we found a formula earlier for T sub one, we can substitute that into our new equation so that we only have one tension force in this equation. So substituting in for T 125 new MS minus M one A minus T sub two equals I A divided by R squared. So if we algebraically solve this for T sub two, then we find that tub two is equal to 25 nuances minus M sub one A minus I A divided by R squared. It's about as far as we can get with that right now. So now let's apply Newton's second law to the box. The hanging mass acting upwards on the box is the tension from two. So T sub two and opposing, that is the weight of the box or M sub two G. And according to Newton's second law, the net sum of the forces is equal to the mass of the box. M sub two multiplied by its net acceleration. And this formula tells us algebraically that T sub two is equal to M sub two G plus M sub two A. So now that we have a formula for T sub two, we can set this equal to the other formula for T sub two that we found a moment ago in terms of the rotational variables. So this tells us that 25 new ones minus M sub one A minus I A divided by R squared is equal to M sub two G plus M sub two A. The purpose of the problem is to find the acceleration. So let's now just take this pro this equation and algebraically solve it for a first by getting all the terms containing a, on their own factoring them out. And then plugging in our variables to solve for a, the first off, we'll get the terms containing a on their own onto one side of the equation. I'm gonna do that by adding to both sides of the equation. The M one A and the I A divided by R squared. So 25 N minus M sub two G equals M sub two A plus M sub one A plus I A divided by R squared. And then factoring up the acceleration is a multiplied by M sub two plus M sub one plus I divided by R squared. So this implies that the acceleration is equal to 25 new ones minus M sub two G all divided by M sub one plus M sub two plus I divided by R squared. As a brief aside, let's talk about that moment of inertia term because we haven't actually established its value yet. Recall the moment of inertia of a disc is equal to one half multiplied by the discs mass multiplied by the square of its radius. The problem gives us both the pulleys mass and its radius. The mass is given to us as 100 50 g or 0.15 kg. And the radius is given as 10 centimeters or 0.1 m we square that put everything into a calculator and we find the moment of inertia to be about 0.75 multiplied by 10 to the power of negative 3 kg meter squared. Now to solve for our acceleration, we'll just take the acceleration equation we found and put in everything that we know that's 25 new ones minus the mass of the box, just 5 g 5 kg multiplied by the acceleration due to gravity and 0.8 m per second squared divided by M 1 4 kg plus M 2 5 kg plus the moment of inertia. 0.75 multiplied by 10 to the power of negative 3 kg meter squared divided by the square of the radius of the pulley, which is 0.1 m. If we put all this into a calculator, then we find the acceleration to be negative 2.64 m per second squared where the negative sign means that the system is moving to the right or downwards. So this is our answer to the first part of the problem. But the next part of the problem asks us to find the tensions in the two different parts of the rope. And earlier in the problem, we basically began the problem by establishing a formula for both of the tension forces. We found t sub one to be equal to 25 new ones. Minus the mass of, of one M sub one multiplied by the acceleration. And we now know the acceleration. So we'll plug in our values, the mass of the slab is 4 kg. And the acceleration we now know is negative 2.64 m per second squared. We put that into a calculator and we find the tension to be negative 36 newtons. So the magnitude is just 36 newtons. We'll do the same thing for the formula we found for T sub two, we found that T sub two is equal to M sub two, the mass of the box multiplied by the acceleration due to gravity plus M sub two multiplied by the total acceleration of the system. So again, we're just gonna plug in our values as 5 kg for M sub two multiplied by 9.81 m per second squared plus the box's mass. Again, 5 kg multiplied by the acceleration negative 2.64 m per second squared. Put that into a calculator and we find the tension to again be 36 new ones at least in that's the magnitude. So the two tension forces are the same. And the multiple choice option that corresponds to what we found is option A. So option A is our answer. That is it for this problem? I hope this video helped you out. And if it did, please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you have a lovely day. Bye bye.

Key Concepts

Newton's Second Law

Tension in a String

Pulley Dynamics

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