Topic: HW 4a #2
Two crewmen pull a boat through a lock, as shown in Figure 5-25. One crewman pulls with
a force of F1 = 145 N at an angle of θ = 32° relative to the forward direction of the raft.
The second crewman, on the opposite side of the lock, pulls at an angle of 45°. With what
force F2 should the second crewman pull so that the net force of the two crewmen is in the
forward direction?
Student A: I’ve tried this problem many different ways and cannot figure it out. i tried
using cos theta and tried to mtch the forces going in that direction using f=ma
and i cannot seem to figure it out. any help?
Student B: example 5-5 page 127 in the book is a good reference for this question
Dr. Man: For this one, it is not a simple cos theta trick,
You should follow problem solving strategy and steps we discussed in the class.
(I will show more example in tomorrow's class)
1 identify and isolate the object of interest
2 identify all forces on the objects acted by other objects (why can you pull your hair
and leave the ground)..
3. pick x – y axis. ( based on acceleration direction for convenience)
4. break up all forces into x-y components
5. Fnet x= m ax ,
Fnet y= m ay (x,y independent)
6. solve for any two unknowns
( from the above two equations)
(Again Math skills here. )
In most cases either ax or ay is zero, if you choose the right x and y directions.
(Or both are zero when the object stay rest or constant velocity)
Student B: You say to solve using F=ma in the x and y directions but how can we do
that without a value for mass? I'm probably thinking about this too much because I kind
of get how the mass wouldn't matter as long as the direction is forward, but I don't know
where to go from here>
Dr. Man: You got the point.
This is a relatively simple question, and mass doesn't really matter for what they are asking.
WHen the direction is forward, the acceleration in the other axis (perpendicular to the river)
is _____?
Remember, once you choose your directions wisely, either ax or ay can be zero.
If ay=0, Total force in y direction =0, no matter what is the mass.
Topic: HW4a #4
A traffic light of mass m (= 18 kg) hangs above the street by two wires, both of which make an
angle (= 40 o) with the vertical.
a.) Show that the tension in both ropes is the same. (Hint: use the horizontal direction.)
(NOT FOR CREDIT, but important for the next parts.)
b.) How much is the tension in each wire?
c.) Find the tension for = 0o.
Student A: This is the one with the traffic light being supported by a rope. You're given the
mass of the light (18kg) and the angle each rope makes with the vertical (40 degrees).
Not sure what I'm doing here; I tried breaking up the components and adding them to
find the tension. Is that what we're supposed to do?
Dr. Man: find all forces, and lable them. Find their horizontal and veritical components.
If you know F, you can find a...
If you know a, you can solve F..
In this case what is acceleration in x direction and y direction for the .traffic light?
As a result, you can set equations that
total force in X direction = ?
total force in y direction = ?
Solve those equations, you will find Tension
Notice that the angle is not 40 degree above horizontal. It is 40 degree away from the vertical
direction, hence 50 degree above horizontal.
Topic: HW4a #5
A crate of mass m (=35 kg) is placed at rest on a (frictionless) inclined plane, which has an
angle (= 50o) above horizontal.
c.) The net force on the crate is:
size:
direction:
Is this greater than or less than your answer from part b? And what is the significance
of your numerical answer at zero degrees? Explain.
Explain why it is impossible to hold up the street light with the wires being perfectly horizontal.
Webassign doesn’t grade your word problems automatically. But we will have those same word
questions in exams.
Student A: If the crate is at rest, that means v=0, therefore, a=0, which makes the net
force also =0. If the net force =0, there wouldn't be a direction, right? I don't know if
I'm thinking about this correctly because there's no such option in the drop down menu.
Dr. Man: it was at rest to start with. v_initial=0
But it has no friction and it is on an incline, what will happen after it is places on the incline?
Any a? which direction?
See lecture notes before working on this problem.
To treat x and y vectors separately is the Key.
If your y axis is in vertical direction, and x axis is horizontal, right, gravity is only
affecting y axis. However, as I mentioned right before the end of the class, in this case,
since the object is likely to move and has acceleration along the incline direction, but
definitely will not move along the direction perpendicular to the incline, it is important
to choose x direction parallel to the incline and to choose the y direction perpendicular
to the incline. Then the gravity has both x and y components. See lecture notes of chapter
three for examples.Also lecture notes on chapter 5 helps.
e):If the crate were kept from sliding by a rope pulling parallel to the incline, what
would the size of the tension be?
Student A: I don’t know how to solve this. How would I go about solving for tension?
Dr. Man: It's still the same. Follow the steps of problem solving.
Find, draw and label all the forces (including the pull), their x and y components...
Set equations for Fnet=ma
in both x and y directions.
If you know "a" in that direction, you can solve forces in that direction. And you know in
order to keep it at rest a=_________?
So the total force along the incline direction should be__________?
BTW, try start HW earlier and ask questions earlier. It's only a few hours before due date
and it is very hard for me to keep replying around midnight. :) Allow 2-3 days!
Topic: HW4a #6
While waiting at the airport for your flight to leave, you observe some of the jets as
they take off. With your watch you find that it takes about 35.5 seconds for a plane to
go from rest to takeoff speed. In addition, you estimate that the distance required is
about 1.1 km.
(a) If the mass of a jet is 1.83 multiplied by 105 kg, what force is needed for takeoff?
Student A: I used the equation Xf-X0= V0*t + 1/2at^2 to find the acceleration then pluged
in what I got to F=ma along with the given mass and it said "Your answer differs
from the correct answer by orders of magnitude." What did I do wrong?
Dr. Man: Pay attention to units etc.
1.1km = ??? m
also the mass of the jet is 1.83*105 kg
Topic: HW4a #7
A 57 kg parent and a 16 kg child meet at the center of an ice rink. They place their hands together and push.
(a) Is the force experienced by the parent more than, less than, or the same as the force experienced by the child?
(b) Is the acceleration of the parent more than, less than, or the same as the acceleration of the child?
Explain.
(c) If the acceleration of the child is 3.1 m/s2 in magnitude, what is the magnitude
of the parent's acceleration?
Student A: How do you find the acceleration with just the mass?
Dr. Man: Look at the leture notes from again.
There was a cute drawing about two people colliding on ice.
Newton's third law tells us that the force A act on B has the same size the force B act on A.
Newton's second law tells us that F=ma , hence a=F/m
Topic: HW 4b #2b-d
A 10 kg crate is at rest on a horizontal table (mus =0.6, muk =0.2). Then various amounts of
horizontal pushing force are applied to it.
Student A: I'm having some trouble finding kinetic friction force, the applied force
is 120N. I've tried finding the the answer by using Uk*N in order to find it, but it ends
up being wrong, and to find the net force I just need N-Fk right?
Dr. Man: recall the experiment I show during the class.
When i pulled the bag using 2, 5, 8, 10 or even 12 Newtons, what happened?
When I pulled with 16 newtons, what happened?
Understand the difference between Kinetic friction and startic friction first.
Understand the upper limit of static friction first.
For the net force, in y direction total force is ?
In x direction the total force is?
If Fnet in y is zero, the total net force is equal to Fnet in X direction.
Read lecture not one more time.
When they ask you size only for forces and accelerations you do not need to include
positive or negative signs. Distinguish normal force and pulling forces, they are
not even in the same direction. again, don't jump and skip steps, form a good habit
and label all forces and solve x and y direction equations separately.
Topic: HW4b #3b
Friction Patch:
A 10 kg crate is sliding at a constant speed of 12 m/s along a frictionless horizontal table,
when it encounters a friction patch (s =1.1, k =0.3). Once the crate enters the friction patch,
a.) What are the sizes of the net force and acceleration of the crate?
size of net force:
size of acceleration:
b.) How fast will the crate be going, when it has travelled 3 m across the friction patch?
Student A: To find the velocity the crate will be going, when it has reached 6m across the
friction path, would I use this equation?:v^2-v0^2= 2a delta x
I am not getting the right answer.
Student B: Well since you are trying to find the final velocity there, you rearrange the equation and
So the equation that you would now have is:
Vf^2 = Vo^2 + 2(a) delta x
And remember acceleration is negative, if you defined the motion direction to be positive.
Dr. Man:
Vf^2 = Vo^2 + 2(a)(delta X), the same as what Student A wrote.
Ok, you found a alreay in part a. yes, Student B is very good to point out that a is negative.
(don't forget to draw yor figures and define your positive direction first.)
you know v0, you know delta X, you don't know t, in order to solve vf, you know which
equation to use.
Student C: how do you find delta X . I am so lost on that part
Dr. Man: Delta x is the displacement! Xf-X0
Once you defined your positive direction, it is straightforward what is the displacement.
(pay attention to the sign. You know distance already).
Topic: HW4b #4
Static and Kinetic Friction on an incline:
A 10 kg crate is placed at rest on an incline plane (s =0.7, k =0.2), whose angle can be adjusted.
a.) The maximum angle of the incline for which the crate will stay at rest is:
b.)-e.) If the angle of the incline were instead the two amounts shown below, please fill in the rest
of the values on the chart below:
angle of incline: component of weight acting parallel to incline: (size only) friction force: (size
only) (indicate static or kinetic):
net force on crate: (size only)
acceleration of crate: (size only)
f.) For the 50o angle, find how long it would take the crate to slide down 5 m along the incline.
g.) Explain why it is NOT possible for an object which is placed at rest on an inclined plane to then
start sliding down with constant speed.
Dr. Man:
This is similar to HW 4a#5, where there is an incline. You should treat and draw your gravity and
normal force the same way as before. Find out the gravity’s contribution along the incline and
perpendicular to the incline.
The only difference is that there is now friction in this problem.
Make sure to identify whether you have static friction or kinetic friction first.
Never use s*N to calculate static friction. Keep in mind that static friction can be any number
between zero and maximum static friction. The actual static friction depends on how much friction
is NEEDED to keep it from sliding. The total force along the incline has to be zero, if stays static.