Need help yet again

[Tanya Back]

Hey Guys! I need ur help with this question, this question is like an incline with a pully..so we have puck which is on an incline surface, and the puck
( mass of 0.5455kg) is attached to a string which is attached to a pully, which also has a string attached to a mass of 0.05 kg. Acceleration is 0.34 m/2^2. The angle is 4.41.

A) Calculate the forces against the puck?

This wut i did-->
Ft= Force of tension
Fa= Forces the against the puck
Fnet= Ft - Fa

Since we don't have force of tesion, i used Fnet = Fg- Ft
Fnet = Fg- Ft
0.05x 0.34= 0.05x 9.8 - Ft
Ft= 0.47265

So then i used in this equation --> Fnet= Ft - Fa
Fnet= Ft - Fa
(0.5455)(0.34)= 0.47265- Fa

Fa= 0.28242N <--- Final answer

B) Should Ff be considered as a major force for your case?

Fn =mgcos4.4
5.3333 N
Fn= Force of gravity prependicular

Fg= 5.3459N ---> 0.5455 x 0.34

Then i did


a^2 + b^2 = c^2
a= Fn
c= Fg
b= force of gravity parallel

I solved for b and i got 0.3711 N
SO then i did Ff = Fa- Fg parallel
0.28242- 0.37115 = -0.08873 N

I got a negative force of firction :surprised ..so now i have no clue wut do to

I am sorre if some of you guys are confused, and plz tell me wut i am doing rite or wrong...this assignment is worth big marks :cry: anyways THANK U in advance.
Tanya

[maverick280857]

Then i did


a^2 + b^2 = c^2
a= Fn
c= Fg
b= force of gravity parallel

I solved for b and i got 0.3711 N
SO then i did Ff = Fa- Fg parallel
0.28242- 0.37115 = -0.08873 N

I got a negative force of firction :surprised ..so now i have no clue wut do to
Tanya

Now I am kinda confused about this step.

First, let me see if I understand the situation correctly....there is a mass m (the puck) on the incline plane, attached to a rope that goes over a pulley attached to the apex of the incline, the other end of which is attached to another mass M.

Now, there are two possibilities. Either the mass M will move downwards taking the puck upwards along the incline or the puck will move down the incline taking M upwards. This depends on the net force on either. If you assume the first possibility, viz that the puck moves down, the force of friction on it will be up the incline. So the forces on it will be mgsin(theta), mgcos(theta), T(tenstion) and friction (f). The net resultant of these forces equals mass of puck * acceleration. For the mass M, the forces on it are T and Mg so T-Mg = Ma. Assuming that the direction of f was correctly chosen, you will get a positive answer. If you're still unconvinced, you should solve the problem considering both possibilites.

Cheers
Vivek