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Question 1 (3 points)A m-5.80 kg box is placed on a ramp as shown, and pulled up it at a constant velocity The angle between the ground and ramp is 0-22.6 degree: C...

Question

Question 1 (3 points)A m-5.80 kg box is placed on a ramp as shown, and pulled up it at a constant velocity The angle between the ground and ramp is 0-22.6 degree: Calculate the magnitude of the applied force that is pulling the box up the ramp. Note: this question is worth 3 marks.A) 40 NB) 66 NC) 60 ND) 44 N

Question 1 (3 points) A m-5.80 kg box is placed on a ramp as shown, and pulled up it at a constant velocity The angle between the ground and ramp is 0-22.6 degree: Calculate the magnitude of the applied force that is pulling the box up the ramp. Note: this question is worth 3 marks. A) 40 N B) 66 N C) 60 N D) 44 N



Answers

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

In this problem, we're given a box sitting on an incline plane of an angle of 30 degrees, and we want to calculate the coefficient of static friction between the box of the incline plane that must exist in order to get this box at rest. So drawing everybody diagram. We have a normal force perpendicular to our surface force of gravity street down. And we have a force of friction going up the ramp to hold this object in place. And if we look at our components of gravity, we have the component that's perpendicular to art inclined and a component that's parallel to our incline. So this component, this, um, adjacent side will be m g co signed data, and this opposite side will be mg signed data. And because object is at rest, we can say M G signed data must be equal to my friction. Now we want to calculate the coefficient of Max static friction in order for this to be held at rest. And so my maximum assuming this 30 degrees is my maximum is going to be my coefficient of static friction. How's my normal force, which is mg co signed data in this case. So we have new times, mg coastline data, my MGs cancel out, and I'm left with signe data divided by co signed data, which is a trick function of tension data gives me my coefficient of static friction so I can do 10 geant of 30 degrees to find my coefficient. Um, 0.58

Let'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'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'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're ready to applying Newton's second law. This is my co ordinate system, By the way, it'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'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'm going to apply Newton's second law on the extraction. In this case, there is an acceleration. So it'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's a new direction and Siegel amassed an assertion. The X we'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'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't write out again, but they're there now. The friction force is still 16.86 news because it'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'm just going to a straight to applying Newton's second law on the extraction so that there is there and I just need to substitute in for my forces. I've a positive aversion force now, a positive way, contribution or gravity contribution, said Sanford Fire, and I subtract my applied force because it'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's what we'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

Their forces on a box on the inclined plane are following. So we have a box here, the force downward. He'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'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'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'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'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'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.

According to given cushions, the box rate is 100 phones. So wait, the blue supposed This is given as 100 phones, not in order to find the force. Since we have to find the fourth magnitude of force needed to pull it to the grand, we can say that the best direction force got this angle is 18 degrees. So this direction forced and wilderness. Since this is W so we can say that is it will do the blue sign, Tito because this is the blue sign and perpendicular direction is w cost Rita So this we can say that hundreds my reply, but signed 18 and 100 multiplied with Zainuddin will multiply its life 100 and does sign Aidan. This is equal to 30 point £9 so force required is 30.9 funds.


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