Reflection Chenbo

What I learned...

We studied vector dynamics and equilibrium during this period. Vector dynamics is about the effect of forces on motion. We use Fnet=ma to solve questions. When there is one object on the flat surface and the other one lean to a side, when m1g > m2gsinθ; system will move in m1 direction. When m2gsinθ > m1g; system will move the other way. Also if the surface has friction, we need to calculate the friction and than use m1g-friction. Then we can get a, if the question asks the tension,we can use T-F1or F2=m1or m2a, if the system move in m1 direction,then a should be negative.
When all the forces that act upon an object are balanced, then the object is said to be in a state of equilibrium. For equilibrium, we use w=mg and use sin, cos and tan to get the tension. Also F left= F right, F up= F down.

Difficulties
For vector dynamics, at the beginning, I will forget to solve for the friction, and also I am not sure to use sin or cos, but it gets better after practice. For equilibrium, I don't know how to draw diagrams at the beginning. It confused me for long time, it's not hard for me now, but I also need to take some time. After drawing a free body diagram, the questions are easy for me.

Problem-solving skills
I am good at vector dynamics, because it's easy for me to draw the diagram and it actually only uses one equation Fnet= ma. We can use it to solve all kinds of vector dynamics questions. 

Reflection Frances

What I Learned…

Recently，i have learnt 2 units which are vector dynamics and equilibrium. In vector dynamics, it shows the force and friction.  vector dynamics along the direction perpendicular to each other two selected vector dynamics called force, the more total  force the use of vector dynamics  by algebra formulas to solve the vector .In the orthogonal decomposition of the force, the purpose of the decomposition of the force, and particularly suitable in the case of the object by a plurality of force, an object F1, F2..., seeking the force F, the various forces in mutually perpendicularde composition, x-axis, y-axis in the x-axis direction of the force component for F1x, F2X,  ... respectively, in the y-axis direction of the force component F1y, F2y, .... then the x-axis direction all force Fx = F1x + F2x +  ... in the direction of the y-axis of the resultant force Fy = F1y + the F2y of + .... all forceset up a joint force with the x-axis angle for θ, then. use orthogonal decomposition method problem solving, the key is how to determine theCartesian coordinate system, the statics, less resolving power and easily decomposed force principle; dynamics, acceleration and vertical acceleration direction to establish coordinate axis, so that Newton's second law of expression: F =ma

 

 

Difficulties...

Sometimes, i will confused with the sin and cos.Before i thought that Fgy equal Fn, actually it is not right.

 

 



When i do this kinds of questions, i don't konw how to slove the tension.

 

 

 

Problem-Solving Skills

 

The first step, the selected object of study and be represented in the form of the particle.
The second step, the force analysis of the selected object.
The third step is the establishment of a Cartesian coordinate system. Generally speaking, in the horizontal plane can be established on the coordinate system, but the best slant build the x-axis direction along the object decline, and then the establishment of the Y-axis.
The fourth step, the analysis of the acceleration direction. Necessary acceleration orthogonal decomposition, in order to do title.
The fifth step is to express aggregate of external forces.
The sixth step, lists the X-direction, and the Y direction on the Newton's second law equation.
Step Seven For other equations, but also to list the equation and then solve.
Step eight ,check it. (Such as the force of the negative undesirable)

 

 

 



 

 

 

 

 

Project: How did the Golden Gate Bridge come to be an engineering classic?

Click here to explore how static equilibrium is applied in this wonderful engineering structure.

Reflection Elaine

What I Learned…

This time we have studied two units: vector dynamics and equilibrium. They are both intriguing topics to explore. Vector dynamics is the physics that’s concerned with the study of forces (magnitude and direction) and their effect on motion. An example would be a person standing in an elevator, where all at the same time the force of gravity, normal force and tension are acting on the system. We focus on the forces that are behind the motion. It’s the opposite of what we learned earlier of the year—kinematics--which deals with motion without reference to its causes (eg. a ball thrown with a velocity and find its maximum height). In vector dynamics, I learned to solve one-dimensional force problems. For example, the elevator problems are concerned with the apparent weight of a person, which is actually the normal force (i.e. scale reading) the floor exerts on that person. Hence, when a person with mass m and weight W is in a rising elevator standing on a scale, the scale would read more than W. This is because the acceleration (a) of the elevator produces a net force (Fnet=ma, Newton’s 2^nd law) and the net force is acting upward, thus adding to the normal force the floor exerts on the person. Another type of one-dimensional force problems involves friction, where a motion is opposed by friction force (Ff=uFN) and we have to remember to subtract friction from the applied force (Fapp) or whatever force that’s acting in the opposite direction of friction when we’re calculating the Fnet in one dimension. For instance, if the mass of a crate and the coefficient of friction (u) are given, the force required to get the sled barely moving can be determined using the formula Ff=uFN=umg, because that required force is friction, which needs to be overcome to cause motion.

Later, we learned about 2-D force problems, taking a step further from 1-D questions. In inclined plane questions, we encountered a more complex force diagram. Let’s say we have a box rested on a 30 degree elevation and we need to find FN. I learned it’s important to note that FN is always perpendicular to the surface from which it’s coming, therefore it’s not always perpendicular to the ground and same as Fg. In fact, FN is the same as the Fgy in this case, the vertical component of Fg. Given Fg and the elevation angle, I can now use cosine to solve for Fgy and by extension, FN.

In addition to vector dynamics, I have also learned about statics, which is the study of forces in objects that are in equilibrium. There are two conditions to guarantee equilibrium. First, the sum of all the forces acting on a system must be zero Newton. This is translational equilibrium, ΣFx=0 N and ΣFy=0 N. Examples include a mass being suspended on two strings, one from left and one from right (ΣFleft=ΣFright and ΣFup=ΣFdown). The second condition for equilibrium, which is the net torque of a system must also be zero Newton meter. This is rotational equilibrium, ΣT=0 Nm. One example is a horizontal beam attached to an upright wall is suspended by a rope from above (ΣTcw=ΣTccw).  

Difficulties...

In the vector dynamics unit, a minor obstacle I ran into concerns with the many body problems. When two different masses connected by a string are being pulled by a string across a surface, there are horizontal forces and vertical forces involved in the free body diagram. I sometimes found it difficult to figure out all the forces in such a system, especially the two tensions in a string that act in opposite directions. It has happened to me that because I only included one of the two tension forces in the force equation, the answer was incorrect.

In the statics unit, on the other hand, my difficulties must have occurred in the ladder problems. It was hard to determine when to apply all of the conditions of equilibrium (ΣFx=0, ΣFy=0, and ΣT=0), but it became clear to me that almost every ladder question requires the application of all three conditions. Another thing about this type of question is that when I was looking for the force the wall exerts on the ladder (FN), I sometimes got confused with the other FN, the one that the ground exerts on the ladder. This time, I think a labelled free body force diagram would be the most helpful.

Problem-Solving Skills

My problem-solving skills have definitely improved as I practiced more from the worksheets and the pre-tests. My strength in problem-solving is drawing a free body diagram. As I read a question, it doesn’t take me long to sketch a force diagram to help me better visualize what I’m given and what I need to solve for. Then after the FBD is drawn, an equation can be easily written out and the question is solved. For instance, a question says: a beam of negligible mass is attached to a wall by a hinge. Attached to the centre of the beam is a 400 N weight. What is the tension in the rope attached at 40 degree at the end of the beam? While scanning the lines, I label the forces related on the diagram or I draw one myself: 

On the other hand, I have my weakness in problem-solving skills, which lies in finding the lever arm of tension from a string. A horizontal beam and a vertical string are not a problem for me, but a slanted beam and a non-perpendicular string are. It was difficult for me to find the lever arm because in such a case, I need to find the one distance that connects the pivot to one end of the tensile string. Further, sometimes we had to decompose tension into x and y (horizontal and vertical) components and multiply these two to get the torque produced by this string. Take this as an example: 

If I am to find the tension in the cable, I need to break the tension into its x and y components, Ty and Force and Tx as Distance, as shown and multiply them together to determine the torque on this tension.

Vector Kinematics

Recall that vectors have both magnitude and direction.

VECTOR KINEMATICS is the study of the motion of objects involving directions. Vector kinematics includes vertically upward thrown projectiles, relative velocity in two dimensions, horizontally launched projectiles, projectiles released at an angle and/or at a height. For the application topic, the focus will be on the case of a projectile released at an angle and at a height.

Reflection Frances

This is my Reflection what I learned about in projectile motion，the difficult problem and how to solve the problems.
What I learned about...

There are 2 parts in projectile motion. Ax is 0 in Horizontal Motion. Ay is constant in Vertical Motion. When it reaches the maximum height, the speed of vertical is always 0. Also , if we start and end at the ground, It will be Vyi = —Vyf. 


Obstacles I encountered...

When I do the questions of projectile motion, I was confused with how to figure out data which are belonging to Horizontal Motion or Vertical Motion. Meanwhile, I was not good at confirming the data of the questions and the direction of graphing.

 
My problem solving skills...

Actually, I just know how to solve the part of projectile motion. I though I must word hard than before. I also do not know the meaning of some words.  

Reflection Chen

This is my refelction about what i learned,the difficults and so on in physic classes. 

what i learned about...

I learned how to solve projectile problems. the horizontal speed of an object it's always constant. Also when it reaches the maxium height, the vertical speed is always 0.I also learned how the wind effect the direction and speed of the airplan, boat and something else. We can use equation to solve these kind of questions Vpg=Vpt+Vtg.

obstacles i encountered...

I find some difficults when i doing vector kinematics. Sometimes i am not sure which direction should the graph be. But then i understand that when a object moving west when we are moving north, we see the object moving south west. It's just like when we stay in a moving car, we see the things go backwards.

My problem solving skills...

I'm good at sloving projectile problems,they are easy for me, but sometimes i will make some small mistakes...My weakness is to make sure the direction of the object athough it's not a problem for me now, but i also need some time to think.