a-box-with-a-mass-of-2-mathrmkg-is-placed-on-an-inclined-plane-that-makes-a-30circ-angle-with-the-ho

Question

A box with a mass of 2 $\mathrm{kg}$ is placed on an inclined plane that makes a $30^{\circ}$ angle with the horizontal. What must the coefficient of static fraction $\left(\mu_{\mathrm{s}}\right)$ between the box and the inclined be in order for the box to be at rest?
(A) 0.5
(B) 0.58
(C) 0.87
(D) 1

Vishal Gupta · Accepted Answer

Hi in the given problem here, this is an inclined plane which is making an angle of 30°.. With the horizontal. There is a block kept over this inclined plane mas of the box. It&#x27;s given as M is equal to two kg force of friction acting on this block will be given by F. Is equal to new. Not times we are. This newest is the coefficient of static friction. So this is new as times the normal reaction. Now, if this is the weight of this box acting vertically downward, It will be having two components 1 normal to this inclined plane. If this angle is 30°,, this will also, with 30° of this component normal to the inclined plane will be MG cost 30° and the component of the weight along the inclined plane which will be having a tendency to bring it down. That is MG signed 30 degree. Now the normal reaction of the inclined plane on the block will be equal to this. Mg cost 30 degree. So four the box to be addressed. The force of friction will be acting here along the inclined plane in upward direction because the tendency of the box is to move downward and we know for so friction always X oppose it to the direction of motion. So for the box to be addressed, the downward force, MG sine 30 degree should be equal to upward force means force of friction means this is you know, times the normal reaction or we can say this is new, not times MG Cost 30°.. So finally canceling this MG. An expression for this coefficient of static friction comes out to be signed 30°,, divided by cost 30° means this is 10 30 degree, whose values one by route three. Or we can say this is zero point 58 Hence here we can see about option B is correct here. Thank you.

Salamat Ali · Answer

their forces on a box on the inclined plane are following. So we have a box here, the force downward. He&#x27;s if wait, scientific data scientist that where the force of friction is outward. So force of friction here and the normal force in this direction and downward. He&#x27;s the the compliment off the wait. So where it is here? The compliment of right here is if the beautiful wait cost that I see then equating the forces on X axis and why direction? Access. Let&#x27;s say this use of our X and this is why So why? Then we can write our forces. Someone forced on the box is equal to zero. We can write. It&#x27;s f normal is a call to if normal is equal to force. Wait times course later course, Vicar. Um then we can write a forces on why Direction? Which our effort Clyde will be cool too. N g scientific minus the forces of friction signs minus a mule mg n g cost data course later, that&#x27;s obstructing the values. Um, we have a mass is a 10 traditional exertions. A 10 sign off 60 sign off 60 year and minus use little 0.2 when deployed by the Mass will be over 10 times 10. I didn&#x27;t times course or for 60 of course. Over. A six see 60. Then they applied force. We get by simplifying this expression is 76.6 Newton. Then, looking into our options, we see Option D is the right answer is the right answer.

Ze-Han Lee · Answer

So in this problem, we have a ramp and ah, the degree of the ram It 30 degrees and we put the block on the rink. Initially, it is arrest. So, uh ah, the friction Constant Mew is a 0.5 in this problem. So the friction acting on this block is ah pointing to this direction. And the magnitude is mg times Coulson lita times me. Okay, Come from you. It&#x27;s not me. Okay? So I&#x27;m James. Course. Aunt Ada is the normal force acting on this block. Okay? And hands me. This gives us the friction. And you&#x27;re gonna see that to this sequel 4.3 time Sam Newton. Yeah. And the the ah, the force that is supporting down lungs along this along along inclines Ah is from is divided from this Ah, gravity. So this is the component allowing this incline. All right, so this this this part of the force has expression after equal. So this is the friction. So Africa omg, I&#x27;m signed either. All right, so this becomes five times and Newton. And as you can see, that the f is greater than friction, right? So, in essence, on the block will accelerate down this ramp with the constable was a constant acceleration. Okay,

Ryan Hood · Answer

let&#x27;s start out by drawing the force diagram where the forces directed down the ramp. And so, if this is my object on the ramp, I&#x27;m a normal force going this way. Gravity going down. I have the applied force going this way. I have a friction force opposing the motion, and I&#x27;m going to break up this gravity into forces along and perpendicular to the ramp. And so this is the angle that they give us. And so this component here is mg time sign of 55 which is the angle they give us. And this is M. G co sign of 55. And now we&#x27;re ready to applying Newton&#x27;s second law. This is my co ordinate system, By the way, it&#x27;s positive X goes down the ramp and positively goes upwards from the ramp. So some of the forces in the white direction equals zero. Since six sellers in the White Direction zero, this implies that the normal force is equal to N g co sign of 55. Since those two forces air, the only force is in the white direction now, uh, blinking in values. Here I get that the normal force is 56.21 Nunes. And then when I played this in to find the friction force I get, that&#x27;s equal to the coefficient of kinetic friction, since the box is moving times normal force, which I should be consistent and indicated by a lower. And as I did here and then when I played the sand, I get 16.86 Nunes. Now I&#x27;m going to apply Newton&#x27;s second law on the extraction. In this case, there is an acceleration. So it&#x27;s NyQuil zeros he called a master acceleration, plugging in all the forces in the extraction. I get positive. Nicholas, positive exponent of the way, grabbed, you should say, minus the friction force, because it&#x27;s a new direction and Siegel amassed an assertion. The X we&#x27;re solving for the acceleration. We just found the fresh and force the applied force is given and the masses given. So everything in this equation is given and we&#x27;re solving for acceleration. And so when we put all that into the calculator, we get that acceleration is positive. 18.34 meters per second squared. So this is acceleration down the ramp that the box experiences in part be the only thing that changes is now the applied forces upwards and the friction forces downwards. So they swap places. The normal force is still this way. Gravity is still down, and it still has the components, the same exact components. I won&#x27;t write out again, but they&#x27;re there now. The friction force is still 16.86 news because it&#x27;s derived purely from the normal force and the coefficient of connect friction. And both those qualities did not change with this new situation here. So this is still the friction force. And now I&#x27;m just going to a straight to applying Newton&#x27;s second law on the extraction so that there is there and I just need to substitute in for my forces. I&#x27;ve a positive aversion force now, a positive way, contribution or gravity contribution, said Sanford Fire, and I subtract my applied force because it&#x27;s now in the negative X direction, and this is equal to mass times acceleration. The X now once again, like last time, everything here is no one except for the acceleration. And that&#x27;s what we&#x27;re solving. Four. So when I do self word I get, the acceleration is equal to negative 2.286 meters per second squared. So the acceleration is up the ramp by this amount and so much less because gravity is working against us now and this

Averell Hause · Answer

so here the best figure to use would be figure 64 in the textbook. Uh, this figure essentially is a very similar system, and we can use the same coordinate system as that figure. So for party to apply Newton&#x27;s second law on the X and Y directions for the Ex direction. We have tea co sign if I minus the fictional force F minus the mass times acceleration in the UAE direction. We have t sign of data plus the normal force minus M g. This is gonna be equaling +20 and we can set the acceleration equaling 20 and the frictional force equaling the maximum static frictional force which would be willing to the coefficient of static friction multiplied by the force. Normal. We know that. Then the force normal is gonna be equaling two mg minus t Sign of data solving, Then for the mass m, we have the mass equaling T over G multiplied by sine If I plus co sign If I divided by the coefficient of static friction, we can say that sold ing for the derivative d m. With respect to d t, we can say this would be equaling two t over G multiplied by co sign of five m vice of m denoting the angle corresponding to the maximum mass that can be pulled. And then we&#x27;re going to say, minus sign of fice of em divided by again the coefficient of static friction. And we&#x27;re setting this equaling 20 And we find that then the tangent of FISA em again, the angle corresponding to the maximum mass. This is going to be equaling to the coefficient of static friction. We know that they can&#x27;t. Then the coefficient of static friction equaling 0.35 And so, if I serve them equaling arc 10 of 0.35 and we have FISA bam equaling 19 degrees. So this would be your answer for party four part B. We&#x27;re going to then plug our value for FISA em into the equation we found for the mass. So we can say that the mass equaling T over G again sign of five plus co sign of fi over, um, use of s and me find that then the mass equaling 340 kilograms. Where again fire here is equaling 19 degrees. So this corresponds to a weight mg, equaling the weight equaling. 340 kilograms multiplied by 9.8 meters per second squared. And this is giving us approximately 3.3 times 10 to the third Newton&#x27;s. This would be our final answer for Part B. That is the end of the solution. Thank you for watching.

Keshav Singh · Answer

for this problem. On the topic of equilibrium and elasticity, we have shown a figure that contains a stationary arrangement of two boxes and three chords. We know the mess of each box, And we know that box a. Is on a ramp at an angle of 30° to the horizontal. We know that the court that is connected to box is parallel to the ramp and frictionless. And we want to calculate firstly the tension in the upper card and secondly the angle that that cord makes with the horizontal. So firstly we know that the tension in the court holding up box B is purely vertical, so there&#x27;s no horizontal component. We need to first find the tension in the cable that is holding A on the ramp. We&#x27;ll call that T. A. So T. A. Is equal to the component of the weight of a parallel to the ram, so that&#x27;s M. A. Times G. Times The Sine of 30°.. And if we substitute our values into this, this is 11 kg Times acceleration due to gravity 9.8 meters per square second times you sign of 30 degrees. And so if we calculate this, we get the tension holding Block A. to B 53 0.9 newtons. Now we need to resolve this vector into its X. And Y. Components and then we can find T. X. And T. Y. So the X. Component of the attention of the upper chord, which will call T. X. Is equal to the X. Component of the cord holding block came place and this is simply 53.9 Newtons times the co sign of 30 degrees. Now detention the white component of the tension in the upper card is equal to the Y. Component of the tension holding block A, which is 53.9 sign of 30 degrees. Plus the tension holding the holding block B. Which is the weight of block B. Seven KGs Times G 9.8 m/km2. And so therefore the tension in the upper card T. Or the magnitude of the tension, is the square root of the sum of squares of the X and Y components. So that&#x27;s 53.9 Co sign of 30°.. All squared. Yes, 53.9 sign of 30 degrees less. seven Kg Times G, which is 9.8. All of this squared and all of this under the square would sign. And so calculating, we get the magnitude of the attention of the upper chord To be 100 and six newton&#x27;s. And so that&#x27;s the magnitude of the tension in the upper cable for the upper chord part B. We want to find the angle that the cord makes to the horizontal. So we know this angle theta By definition, is the Act 10 of T. Y by T. X. So if we substitute T. Y and T. X. From above into the situation, we get this angle data to be 64 degrees.

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