query 251

course phy201

Query 25.1025. More Forces

*********************************************

Question: `q001. Note that this assignment contains 5 questions.

. A pendulum consists of a 150 g mass suspended from a light string. Another light string is attached to the mass, which is then

pulled back from its equilibrium position by that string until the first string makes an angle of 15 degrees with vertical. The

second string remains horizontal.

Let the x axis be horizontal and the y axis vertical. Assume that the mass is pulled in the positive x direction. If T stands for the

tension in the pendulum string, then in terms of the variable T what are the x and y components of the tension?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x= 150g cos ( 15deg)=144.89

y= 150g sin ( 15deg)=38.8

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The pendulum string makes an angle of 15 degrees with vertical. Since we have assumed that the pendulum is pulled in the positive x

direction, the direction of the tension in the string will be upward and to the left at an angle of 15 degrees with vertical. The

tension force will therefore be directed at 90 degrees + 15 degrees = 105 degrees as measured counterclockwise from the positive x

axis.

The tension will therefore have x component T cos(105 degrees) and y component T sin(105 degrees).

STUDENT QUESTION:

why did you add 90deg to 15deg, i knew that x should have been Tcos(15) and y Tsin(15), i wasn't sure about the 90deg

INSTRUCTOR RESPONSE

15 deg is the angle with the y axis. If you use T cos(theta) and T sin(theta) then the angles must be measured counterclockwise from

the positive x axis.

The pendulum is displaced in the positive x direction. So relative to the position of the pendulum mass, the string pulls up and to the

left--into the second quadrant--at an angle of 90 degrees + 15 degrees = 105 degrees.

Starting from the positive x axis you would have to rotate through 90 degrees to get to the y axis, then through the 15 degree angle

the string makes with the y axis.

STUDENT COMMENT: Ok that makes sense. Its like the vertical pendelum is just shifted so the pendelum tension will be like a vector.

INSTRUCTOR RESPONSE Right. The tension exerts a force, and forces are characterized by magnitude and direction so they can be

represented by vectors.

STUDENT QUESTION

I didn’t know that I had to add 90 degrees to the y-axis angle, but that explains my confusion earlier.

INSTRUCTOR RESPONSE

If you use the circular definition of angles, as is done in the given solution, the y axis is at the 90 degree position and an angle of

15 degrees with the y axis is at 90 deg + 15 deg or 90 deg - 15 deg; in this case the angle is 90 deg + 15 deg = 105 deg.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

So we must add 15 to 90 = 105

x= t cos (105degrees)

y= t sin ( 105degrees)

------------------------------------------------

Self-critique rating #$&*ent:3

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

*********************************************

Question: `q002. Continuing the preceding problem, we see that we have a vertical force of T sin(105 deg) from the tension. What

other vertical force(s) act on the mass? What is the magnitude and direction of each of these forces? What therefore must be the

magnitude of T sin(105 deg).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The other vertical forces acting on the 150g mass are the forces of gravity.

F= m * a

F= .15kg * 9.8m/s^2 = 1.47N

x= 1.47 cos (15)=

y= 1.47sin (15) =

?not sure if I should do this???

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The only other vertical force acting on the mass will be the gravitational force, which is .150 kg * 9.8 meters/second ^ 2 = 1.47

Newtons. The direction of this force is vertically downward.

Since the mass is in equilibrium, i.e., not accelerating, the net force in the y direction must be zero. Thus

T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons.

STUDENT QUESTION

Ok. So what I basically do is set the gravitational force equal to that of T sin(105). And the 1.47 Newtons of gravitational force

now becomes the magnitude of the x component….???

INSTRUCTOR RESPONSE

The gravitational force remains what it is--just the gravitational force.

We assume the gravitational force to be in the downward vertical direction.

Since the mass is assumed to be in equilibrium, with the only other force in the y direction being the y component of the string

tension (which is T sin(105 deg) ), it follows that these two forces are in equilibrium so that

T sin(105 deg) = 1.47 N.

Nothing in the solution to this question addresses the x component of the force, but that's coming up in the next question.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons.

??????????????????????So this means that the only force acting on the object is the force of gravity, since there is no acceleration y

is = to zero, correct???????????????????????

The reason this quantity is zero is that the net force in the y direction is zero. We know this because the object accelerates only in the x direction (y acceleration is negligible).

In the y direction we have both the tension and the force of gravity, which must be equal and opposite.

------------------------------------------------

Self-critique rating #$&*ent:3

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

*********************************************

Question: `q003. Continuing the preceding two problems, what therefore must be the tension T, and how much tension is there in the

horizontal string which is holding the pendulum back?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x= 1.47 cos (15)=

y= 1.47sin (15) = .380

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons.

Thus the horizontal component of the tension will be T cos(105 deg) = 1.52 Newtons * cos(105 deg) = 1.52 Newtons * (-.26) = -.39

Newtons, approximately.

Since the mass is in equilibrium, the net force in the x direction must be zero. The only forces acting in the x direction are the x

component of the tension, to which we just found to be -.39 Newtons, and the tension in the second string, which for the moment will

call T2. Thus

T2 + (-.39 N) = 0 and T2 = .39 N.

That is, the tension in the second string is .39 Newtons.

STUDENT COMMENT: I'm really confused now. If we started out with a .15 kg mass that is equal to 1.47 Newtons. How did we create more

weight to get 1.52 Newtons? Is the horizontal string not helping support the weight or is it puling on the weight adding more force?

INSTRUCTOR RESPONSE: A horizontal force has no vertical component and cannot help to support an object against a vertical force.

The vertical component of the tension is what supports the weight, so the tension has to be a bit greater than the weight.

The tension in the string is resisting the downward weight vector as well as the horizontal pull, so by the Pythagorean Theorem it

must be greater than either.

STUDENT COMMENT: Thats odd how that works that the tension is negative

INSTRUCTOR RESPONSE: The tension isn't negative, but in this case, where the displacement from equilibrium is positive, the string

pulls back (upward and to the left) so the horizontal component of the tension is negative.

STUDENT QUESTION

Is where I seemed to get off is I assumed I could set the X component equal to the force in Newtons found from the Y direction

(.15kg*9.8m/s^2).

From what I can tell now, the tension in X is equal to the X component being set equal to the force in the Y direction * the

displacement from the Y axis…I THINK that is a way of looking at it.

INSTRUCTOR COMMENT:

Here's an overview of what we know and how we use it. The given solution will fill in the details at the end.

Let T be the unknown magnitude of the tension vector.

We know that T is at 105 degrees.

The components of T are therefore

T_y = T sin(105 deg) and

T_x = T cos(105 deg).

The y component of the tension is what supports the mass of the pendulum in opposition to the force exerted on it by gravity.

Setting T sin(105 deg) equal to m g we find T, as shown in the give solution.

Then we can find the x component T_x, as shown in the given solution.

STUDENT QUESTION

I’m not sure where the -.26 came from.

INSTRUCTOR RESPONSE

-.26 is roughly equal to the cosine of 105 degrees

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

The solution says....... If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons.

???????????????????shouldnt T= 1.47 Newtons / (cos(15degrees)) = 1.47N/ .97 = 1.52N in the horizontal direction??????????????????

cos(105)= -.26

then we mulitply the tension from the horizontal direction 1.52 * -.26= -.39

So vertically gravity is the only force action on the weight which is creating tension on the string vertically so Tsin(105)= 1.47,

vertically. I think I understand this.

Horizontally we divide the tension force from gravity 1.47/ by the cos(15degrees) to get the tension acting on the horizontal from the

vertical force, then we must find the tension horizontally by cos(105)= -.26 and mulitply that by the 1.52N acting from the vertical

positon. So it is like the area almost.????????????????????????????????????????????????

This is kind of confusing

You've just about got it sorted out. Hopefully the following will help:

The tension is a vector of magnitude T at angle 105 degrees. Its components are T_x = T * cos(105 deg) and T_y = T * cos(105 deg).

Its vertical component must be 1.47 Newtons in order to keep the mass from accelerating in the y direction (i.e., to keep it from falling). In order to have 1.47 Newtons in the vertical direction the tension must be 1.52 N (shown in preceding problem).

Once we know this it's easy to find T_x, which is just T * cos(105 deg) = 1.52 N * cos(105 deg) = =.39 N, approx..

------------------------------------------------

Self-critique rating #$&*ent:2

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

*********************************************

Question: `q004. If a 2 kg pendulum is held back at an angle of 20 degrees from vertical by a horizontal force, what is the magnitude

of that horizontal force?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

magnitude of the horizontal would be 90+20= 110

x= fhorizontal cos (110)

y = fhoriz sin(110)

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

At the 20 degree angle the tension in the pendulum string will have a vertical component equal and opposite to the force exerted by

gravity. The tension with therefore have a horizontal component. To achieve equilibrium by exerting the horizontal force, this

horizontal force must balance the horizontal component of the tension.

We therefore begin by letting T stand for the tension in the pendulum string. We also assumed that the pendulum is displaced in the

positive x, so that the direction of the string as measured counterclockwise from the positive x axis will be 90 degrees + 20 degrees

= 110 degrees. Thus the x component of the tension will be T cos(110 deg) and the y component of the tension will be T sin(110 deg).

The weight of the 2 kg pendulum is 2 kg * 9.8 meters/second ^ 2 = 19.6 Newtons, directed in the negative vertical direction. Since

the pendulum are in equilibrium, the net vertical force is zero:

T sin(110 deg) + (-19.6 N) = 0

This equation is easily solved for the tension: T = 19.6 N / (sin(110 deg) ) = 19.6 N / (.94) = 20.8 Newtons, approximately.

The horizontal component of the tension is therefore T cos(110 deg) = 20.8 N * cos(110 deg) = 20.8 N * (-.34) = -7 N, approx.. To

achieve equilibrium, the additional horizontal force needed will be + 7 Newtons.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

ok I got the

x= T cos(110)

y= T sin(110)

F= 2kg * 9.8m/s^2= 19.6N

NEt vertical force = 0

19.6N/ (sin(110)) = 19.6/ .94 = 20.8N

?????????????????????????So is this the force vertically that acts on the horizontal????????????????????????????

The 20.8 N is the tension in the string, which acts along the direction of the string, with components in both the x and y directions.

The horizontal component of this tension is 20.8 N * cos(110 deg).

Then the horizontal is

cos(110)= -.34

20.8N *-.34 = - 7.11N

------------------------------------------------

Self-critique rating #$&*ent:2

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

*********************************************

Question: `q005. The 2 kg pendulum in the previous exercise is again pulled back to an angle of 20 degrees with vertical. This time it is held in that position by a chain of negligible mass which makes an angle of 40 degrees above horizontal.

Describe your sketch of the forces acting on the mass of the pendulum.

What must be the tension in the chain?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

90 + 40 = 130degrees

T sin (130)

T cos (130)

confidence rating #$&*

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The weight of the pendulum is partially supported by the tension in the chain. Thus the tension in the pendulum string is not the

same as before. The horizontal component of the tension in the chain will be equal and opposite to the horizontal component of the

tension in the pendulum string.

Your picture should show the weight vector acting straight downward, the tension in the pendulum string acting upward and to the left

at an angle of 20 degrees to vertical and the tension in the chain should act upward into the right at an angle of 40 degrees above

horizontal. The lengths of the vectors should be adjusted so that the horizontal components of the two tensions are equal and

opposite, and so that the sum of the vertical components of the two tensions is equal of opposite to the weight vector.

Since both tensions are unknown we will let T1 stand for the tension in the pendulum and T2 for the tension in the chain. Then T1, as

in the preceding problem, acts at an angle of 110 degrees as measured counterclockwise from the positive x axis, and T2 acts at an

angle of 40 degrees.

At this point whether or not we know where we are going, we should realize that we need to break everything into x and y components.

It is advisable to put this information into a table something like the following:

x comp y comp

T1 T1 * cos(110 deg) T1 * sin(110 deg) in

T2 T2 * cos(40 deg) T2 * sin(40 deg)

Weight 0 &nb sp; -19.6 N

The pendulum is held in equilibrium, so the sum of all the x components must be 0, as must the sum of all y components. We thus obtain

the two equations

T1 * cos(110 deg) + T2 * cos(40 deg) = 0 and

T1 * sin(110 deg) + T2 * sin(40 deg) - 19.6 N = 0.

The values of the sines and cosines can be substituted into the equations obtain the equations

-.33 T1 + .77 T2 = 0

.95 T1 + .64 T2 - 19.6 N = 0.

We solve these two simultaneous equations for T1 and T2 using one of the usual methods. Here we will solve using the method of

substitution.

If we solve the first equation for T1 in terms of T2 we obtain T1 = .77 T2 / .33 = 2.3 T2. Substituting 2.3 T2 for T1 in the second

equation we obtain

.95 * 2.3 T2 + .64 T2 - 19.6 N = 0, which we easily rearrange to obtain

2.18 T2 + .64 T2 = 19.6 Newtons, or

2.82 T2 = 19.6 N, which has solution

T2 = 19.6 Newtons/2.82 = 6.9 N, approximately.

Since T1 = 2.3 T2, we have T1 = 2.3 * 6.9 N = 15.9 N, approximately.

Thus the pendulum string has tension approximately 15.9 Newtons and the chain the tension of approximately 6.9 Newtons.

STUDENT QUESTION:

I thought we already found the tension for the original string so we just have to solve for the T components in the

chain. How come the tension found earlier is not the same one we use here?

INSTRUCTOR RESPONSE:

Previously the 'pullback' used a horizontal string, which exerted no force in the vertical direction.

The string is now at a 40 degree angle, so any tension must have both horizontal and vertical components.

Since we need a horizontal component to pull the mass back, there will be a nonzero vertical component. This will have the effect of

reducing the tension in the pendulum.

STUDENT QUESTION

I read through the Given Solution, and I think I understand. I’m just not very good at recognizing solutions where I have to use

substitutions.

But one question…. When you were solving for T1 * cos(110) + T2 * cos(40) = 0, I solution for T1 = -2.3 * T2. Why did you change the

solution from negative to positive?? Does it have something to do with the tensions being equal and opposite, or what?

INSTRUCTOR RESPONSE

The given solution didn't change any signs. The equation T1 * cos(110) + T2 * cos(40) = 0, solved for T1, gives you T1 = + 2.3 * T2.

You just have to be careful about the signs of your trig functions.

T1 cos(110 deg) + T2 cos(40 deg) = 0, so that T1 = -T2 * cos(40 deg) / cos(110 deg).

cos(110 deg) is negative so the relationship is T1 = -T2 * (.77 / (-.33) ) = +2.3 T2. The tension T1 is not -2.3 T2.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

ok

For the 20 degree T1=

cos (110)= -.34

sin(110) = .94

For the forty degrees T2=

cos(40) = .76

sin(40) = .64

?????????we do not have to add to 90 to the 40 b/c it is with the Horizontal correct???????????????????????????????????

Right. That's 40 deg to the right and above horizontal, which is in the first quadrant.

So the equation are

-.33 T1 + .77T2= 0

.95 T1 + .64 T2 - 19.6N = 0

we solve for T1 T1= .77T2/.33= 2.33T2

Then substitute T1 in the second equation to sole for T2

.95 (2.33T2) + .64 T2 -19.6N = 0

2.21 t2 + .64T2 -19.6N = 0

2.85T2- 19.6N = 0

2.85 T2 = 19.6N

T2= 6.87N

Then multiply T1 * T2 = 2.6 * 6.9N = 15.9N

So the pndulum string has forces of both horizonal and vertical forces at 15.9N and the chain has a force of only the 6.9N

horizontally.

The chain tension exerts a force at 40 degrees, not a purely horizontal force.

???????????????????????????So adding the x forces together times the respective tensions and they equal zero gives us the first

equation, then adding the y components together times the respective tension along with the force of gravity gives us the second

equation.??????????????????????????????

That is a good statement.

------------------------------------------------

Self-critique rating #$&*ent:2

&#Good responses. See my notes and let me know if you have questions. &#