(II) Two blocks made of different materials connected together...

Question

(II) Two blocks made of different materials connected together by a thin cord slide down a ramp inclined at an angle θ to the horizontal, Fig. 5–40 (block B is above block A). The masses of the blocks are m_A and m_B, and the coefficients of friction are μ_A and μ_B . If m_A = m_B=4.0kg, and μ_A = 0.20 and μ_B = 0.30, determine
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(a) the acceleration of the blocks and

(a) the acceleration of the blocks and

Accepted Answer

Hello, let's go through this practice problem. A wooden block labeled W in the diagram and a metallic block labeled M are tied together by a lightweight cord and slid down an inclined plane at an angle of theta equals 30 degrees to the horizontal. The masses of the blocks are M sub W equals M sub M equals 5 kg. The coefficients of friction for the wooden and metallic blocks are 0.25 and 0.35 respectively. Calculate the acceleration of the system as the blocks slide down the incline option. A 1.5 m per second squared. B 2.4 m per second squared, C 3.3 m per second squared and D 4.9 m per second squared. So because this is a problem that involves forces and analyzing things such as friction and mass and angles on an inclined plane, we're going to want to begin this problem by drawing out our forces on the two blocks. So to start us off, let's make things a little easier on ourselves because both blocks are resting on an inclined plane. Let's set up an inclined coordinate system where the X axis is positioned along the inclined plane itself and the Y axis is perpendicular to that. So let's draw the forces acting on the two blocks. Both blocks will have a force from weight downward on them. So for mass M, there is a weight of W sub M and for block, W there's a downwards weight of W sub W both blocks will have normal forces that are perpendicular to the surface of the plane. So block, W as a normal force of N sub W and block M is a normal force of N sub M, there are normal force. There are frictional forces opposing the motion of the two blocks. So acting in the opposite direction of the downward force. So there's a frictional force of F sub KW acting in block W and a frictional force of F sub KM acting on block M. Since both blocks are attached by a string, there's also a tension force to consider. So there's a tension force acting downwards from block W and acting upwards from block M. I'm not going to label the two tension forces differently because since the two blocks are connected to each other by the same string, those tension forces are going to be equal when you consider Newton's third law. And recall that Newton's third law states that if object A is exerting a force on object B, then object B is exerting a force of the same magnitude on object A. So if the two blocks are connected to each other by a chord and we're ignoring the mass of the court itself. So there are no other forces there to consider. Then we can assume that the tension forces are equal. Anyways, that is it to consider for the forces acting on the two blocks. Now let's begin writing out some force equations to see how this all fits in. Let's start by considering the horizontal forces acting on the wooden block. So F net W so because we have an inclined coordinate system, that means that there is now a horizontal and vertical component of the weights. So pulling the block downwards is going to be the horizontal component of that weight. So M sub W multiplied by G multiplied by the sine of the angle, theta 30 degrees pointing in the opposite direction upwards up the ramp is the frictional force acting on W. So F sub W and then pulling it back down the ramp is the force from the tension. Now, let's write a similar equation for the net horizontal for, for the horizontal forces on the metallic block. So once again, we have to consider the horizontal component for the weight. So it's M sub mg multiplied by the sine of theta minus the frictional force on block M minus the tension. In this case, we're subtracting tension since for the metallic block tension points up the ramp. Now considering the vertical forces, if you look at the diagram, you'll see that the only vertical forces acting on the two blocks are the normal force and the vertical component for the weight. And we can actually just write one equation that'll account for both blocks here. Because of the fact that both blocks have the same mass, the normal force for each block is equal to the component of the gravitational force that's perpendicular to the incline, which if the two weights are the same doesn't change. So the normal force for each block is equal to MG, the weight of the block multiplied by the cosine of the angle theta. So the mass of the block is 5 kg for each of them multiplied by the gravitational acceleration 9.8 m per second squared multiplied by the cosine of the angle theta which is 30 degrees, put this into a calculator. And we find that the normal force on either block is 42.435 newtons. So remember the problem was asking us to find the acceleration of the whole system. Let's first find the total force of the entire system. So F net total and we can find the total force in the system by combining the net force on block W plus the net force on block M. So we're adding M sub WG sine theta minus F sub W plus T. And all this is being added to the net forces and block M which are M sub MG sign theta minus F sub M minus T and right off the bat notice that there are some ways we can simplify this formula since the two TS have opposite signs, they can cancel each other out. Since we have two terms containing MWG sign data, we can combine those into a single term. That part can be written as two MG sine theta. Recalling the fact that the two masses are the same and then we have the two frictional terms which can be rewritten when we consider how frictional forces are. Recall that frictional force is equal to the coefficient of friction multiplied by the normal force. So the frictional force for block W can be rewritten as the coefficient of friction for block W multiplied by N, then we add or subtract the frictional force on block M. So the coefficient of friction for block M multiplied by the normal force. And notice that in these two terms containing the normal force, we can factor out the normal force. So this whole equation can be rewritten as two MG sine theta minus me in parentheses. Mu sub W plus mu sub M multiplied by N. So now to find the total force, we plug in the values that were given to us in the problem. That's two multiplied by the mass of 5 kg multiplied by the S or multiplied by the acceleration due to gravity of 9.81 m per second squared multiplied by the sine of the angle 30 degrees minus the sum of the coefficient of friction. So minus two point point 25 plus 0.35 multiplied by the normal force of 42.435 newtons. And putting all this to a calculator, the total force turns out to be 23.539 nuance. OK. So we've got the total force on the system, but the problem is asking for us to find the total acceleration. So this is where we're again going to go back to New and second law because recall that according new and second law, the acceleration of an object is equal to the total uh net forces acting on the object divided by the mass of the system. Well, in this case, we just found the force on the system. It's 23.539 newtons, then we divide it by the mass of the whole system which we just add the two masses of the blocks. Both are 5 kg. So the total mass is five plus five, which is 10 kg, put this into a calculator. And we find the acceleration to be approximately 2.4 m per second squared. And that is it for this problem as this corresponds to option B. So option B is our answer and that is it. I hope this video helped you out and 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.

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