Calculate the forces against the puck?

[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 ..so now i have no clue wut do to

I am sorre if some of you guys are confused, and please tell me wut i am doing rite or wrong...this assignment is worth big marks 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 ..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

 

[wais]

Hi Tanya, great job on starting the problem! Let's break down the forces on the puck and see if we can get a better understanding of what's happening.

First, let's draw a diagram of the situation to help us visualize it better. We have a puck on an inclined surface, attached to a string that goes over a pulley and is connected to a mass of 0.05 kg. The angle of the incline is 4.41 degrees and the acceleration is 0.34 m/s^2.

We can break down the forces acting on the puck into three components: the force of tension (Ft) from the string, the force of gravity (Fg), and the force of friction (Ff).

1. Force of Tension (Ft): This is the force pulling the puck up the incline. We can calculate this using the equation Ft = ma, where m is the mass of the puck and a is the acceleration. Plugging in the values, we get Ft = (0.5455 kg)(0.34 m/s^2) = 0.1854 N.

2. Force of Gravity (Fg): This is the force pulling the puck down the incline. We can calculate this using the equation Fg = mg, where m is the mass of the puck and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get Fg = (0.5455 kg)(9.8 m/s^2) = 5.3459 N.

3. Force of Friction (Ff): This is the force acting in the opposite direction of motion, slowing down the puck as it moves up the incline. We can calculate this using the equation Ff = μN, where μ is the coefficient of friction and N is the normal force. The normal force is the component of the force of gravity that is perpendicular to the surface of the incline. We can calculate this using the equation N = mgcosθ, where θ is the angle of the incline. Plugging in the values, we get N = (0.5455 kg)(9.8 m/s^2)cos(4.41) = 5.3333 N. Now we can calculate Ff using the coefficient of friction given to us. We'll assume a coefficient of friction of 0.2 for this problem