- #1

- 11

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Fnet = ma

Fw = mg

Ftx = Ftcos37

Fty = Ftsin37

So first I tried solving for the Fw using Fw = mg. After finding that I assumed that Fw = Ft and used it to solve for Ftx, but then i'm stuck and have no idea what i'm doing.

- Thread starter aznboi855
- Start date

- #1

- 11

- 0

Fnet = ma

Fw = mg

Ftx = Ftcos37

Fty = Ftsin37

So first I tried solving for the Fw using Fw = mg. After finding that I assumed that Fw = Ft and used it to solve for Ftx, but then i'm stuck and have no idea what i'm doing.

- #2

- 56

- 0

From such frame you can see that the body you got is hanging from the ceiling forming a certain angle (it is at rest!). Now, only two real forces on the body are gravitational force and string tension. But, according to 1. Newton's net force on the body must be zero for body to be at rest. So we introduce one fictive force (inertial force). Its magnitude is [tex]F_{i}=ma[/tex], where [tex]a[/tex] is acceleration of the cart. This force is always pointing in opposite direction of the cart acceleration. (Remember: this force actually doesn't exist!)

Now, only thing you need to do is to represent each of the forces you have (gravitational, tension and fictive force) along the x and y axis. Net forces along those axes must be zero.

So, you have these equations ([tex]T[/tex] is the tension force and [tex]\alpha[/tex] is the angle):

[tex]T \sin \alpha - ma = 0[/tex] (x axis, right is taken as positive direction)

[tex]T \cos \alpha - mg = 0[/tex] (y axis, up taken as the positive direction)

From there, you can easily obtain magnitude of the acceleration and tension force.

- #3

- 11

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Ok, so I understand that when the cart is accelerating, the hanging mass is at rest and what you propose make sense. So for this case, is the Ft equals to the Fw since the object is at rest? And I don't get this fictive force thing =|.

Btw, the answers are:

a) 13 m/s/s to the right

b) 8.3n

EDIT: Sorry I read the question wrong, the weight is only .50 kg!!

Btw, the answers are:

a) 13 m/s/s to the right

b) 8.3n

EDIT: Sorry I read the question wrong, the weight is only .50 kg!!

Last edited:

- #4

- 56

- 0

Then you look at the net forces along each of the axes:

[tex]ma_{x}=T \sin \alpha[/tex] ([tex]a_{x}[/tex] is acceleration along x axis)

[tex]ma_{y}=T \cos \alpha - mg[/tex] ([tex]a_{y}=0[/tex] is acceleration along y axis)

From these two equations you get

[tex]a_{x}=g \tan \alpha[/tex]

[tex]T=\frac{mg}{\cos \alpha}[/tex]

where [tex]\alpha = 90^{\circ}-37^{\circ}=53^{\circ}[/tex] (I messed up the angle a bit because I didn't read the text of the problem well, sorry.) [tex]a_{x}[/tex] is the acceleration you are looking for (the acceleration of the body looking from the inertial frame of the Earth, not from the cart). Acceleration of the body is equal to the acceleration of the cart (again, if you look from outside the cart).

Sorry if I confused you with too much theory, but I think it is important to know what's going on.

- #5

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Thank you! I understand now.

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