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Two ropes are attached t0 The rope with a 50 - force points along direction 40e from the positive x-axis 40 kg object Olher robe with 50 N force points along direc...

Question

Two ropes are attached t0 The rope with a 50 - force points along direction 40e from the positive x-axis 40 kg object Olher robe with 50 N force points along direction 70"_ What is the ueceleration of the object? A) 14.8 m/s? B) 16.32 m/s? 40 ms? D) 7.515 mls? 6.8 ms?

Two ropes are attached t0 The rope with a 50 - force points along direction 40e from the positive x-axis 40 kg object Olher robe with 50 N force points along direction 70"_ What is the ueceleration of the object? A) 14.8 m/s? B) 16.32 m/s? 40 ms? D) 7.515 mls? 6.8 ms?



Answers

A 20 -kg crate hangs at the end of a long rope. Find its acceleration (magnitude and direction) when the tension in the rope is ( a) 250 $\mathrm{N},(b) 150 \mathrm{~N},(c)$ zero, (d) $196 \mathrm{~N}$.

Oh in that problem we have an object whose mass is five kg. So here This is five kg and it's stranded in a rope. And the acceleration of that mass is 4.3 in the upward direction. So opening three m per second squared in the positive direction. So, this object here is under the effect of two different forces. First, the force of weight which is always pointing downward in the direction of gravity, and the other force is the tension in that rope here. So, from Newton's law, we have that the sum of forces acting on a certain object is equal to the mass, multiplied by the acceleration. So from here the mass of this object here Is five and the acceleration is opening three. This gives us a total resultant force of 1.5 newtons. So now we know that the sum of forces Is equal to 1.5 Newtons. So now, what are the forces acting on that object? As we said, we have the force of weight plus the force of tension These two added together should end up in 1.5 newtons pointing upward here, we will denote the positive signed to indicate an upward direction, and vice versa downward means negative. So here the force of weight would take a negative sign. While the force of tension will take a positive sign. Let's calculate this force of weight explicitly. We know that the weight of the object is equal to the mass of the object multiplied by the gravitational exploration. And here let's give this a negative sign. And then we have a positive sign for the force of tension. And that's again equal to 1.5. Noten's. Now let's substitute The masses five kg multiplied by the gravitational exploration, which is 9.81 plus the force of tension Equal to 1.5. This term here is equal to 49 25. And let's do not forget the negative sign plus the force of tension. This is equal to 1.5. From here. We can take this term here to the other side with a positive sign And we will have that. The force of tension is equal to 1.5 plus 49 went oh five, which is equal to 50 0.5 five. No tints or approximately 51 Noten's. And that's the value for the tension in the court. Mm.

In this problem. We have a mass supported by a rope intention, which is tied to two other ropes at a nod, were given one of the tensions and were asked to find the two other tensions and the mass. To begin with, we can look at a free body diagram of the mass. We see that there's only two forces acting on it and the masses in equilibrium so we can state that are t one equals two mg. And then we can look at the not in equilibrium. And we can some the forces in the X and the Y direction and set 20 to solve for all three quantities in the extraction, we have the force. We have 100 cosine 50 minus 83. And in the why we have a 100 sign 50 minus m g. All right, and we'll get r T three equals to 100 co sign 50 and we'll solve the why equation to find our mass equals 100 sign 50 over G. And then we can plug that back into the equation for key one. And we find t one equals 100 sign 50 your grades okay. And if we crunch our numbers here, we will get. She won equal to 80. Newton's she three equal to 60 Newton's and our Mass equal to seven point eight kilograms. 22 significant figures.

This question covers the concept of the *** conservation and the concept of the centripetal acceleration. If we consider The gravitational potential energy at horizontal level is zero, which means the equals zero at this point. So from energy conservation for party, we can write the energy at initially that is uh zero and it is equal to the final energy and the final gravitational potential energy is minus mg times the length of the string. Help plus the kind of technology that is half of M. V. Squared. Let's say this period when the string is vertical is B. And from this V equals square root off two G times the length of the string. Now we can substitute the value to find the speed of the mass when the string is vertical and that is a Cuban, two square root of two and two G's 10 m for second square into the length and the length is two m. Ah the speed equals 6.32 m/s. Okay, now part B. The centripetal acceleration at that instant is we square upon our and we square is 6.32 m/s square. And the radius is the length of the string and that is two m. The centripetal acceleration is 20 m/s squared for part C. When the mass is vertical, the forces on the masses, the weight acting vertically downward and that is MG. An attention which is acting vertically airport steep and the acceleration of the objectives along the port direction. And that is a for party, the net radial force equals uh the centripetal force that is mass into the acceleration. But we can try potential minus MG equals the mass times the acceleration. Or the tension is the mars into the acceleration. Plus she If you substitute the value, the tension in the string. Is the masses five Kg into the acceleration. That is 20 plus the gravitational acceleration that is 10 or that engine is in Cuba. And to 1 50 Newton's

So this probably again hung a light using two ropes attached to our house. In this case, we don't know what the mass of the light is, but we know that the tension in this rope is 100 Newtons and it makes an angle of 50 degrees with the positive X axis. And this rope is court is horizontal and it has some unknown force in it. So what we're looking for in this problem is M and F one again when you could just write force balances in the UAE direction. We know that, um, 100 Newtons times sign of 50 degrees so that why component of that force must equal mg. So we can then saw for M. And we get that it is 7.8 kilograms. Force balances in the wide to read in the ex direction. Now we know that the X component here is 100 co signed 50 degrees, and that has to equal the X component or the total value of F one. And so you can plug in our numbers and we get F one equal 64.3. Newton's


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