Help with Physics w/ Cal 1 & Projectile Motion

[sb_4000]

for the block hanging
F_x= 0
f_y = T +(-g)

 

[Pyrrhus]

Almost there.

are you familiar with vector components?

By the way, when it means Force of Gravity (weight) it means mass x gravity.

 

[Pyrrhus]

for the block hanging
F_x= 0
f_y = T +(-g)

Read my above post, also you're forgetting something from the formula.

By the way, are you familiar with sign conventions?

 

[sb_4000]

what are the sign conventions?

 

[sb_4000]

what am I missing from the formula?

cos(theta) and sin(theta)?

 

[Pyrrhus]

In your system, when you draw the free body diagram, you must choose which way is positive in the coordinate system. The standard is right and up positive, and left and down negative, so any vectors pointing left or down will appear as negative in the scalar equations, while the vectors pointing right and up will appear as positive. This is very important.

 

[Pyrrhus]

what am I missing from the formula?

cos(theta) and sin(theta)?

On the hanging block , it's not -g, it's mg ot -mg depending on your sign convention, and you're forgetting may.

On your analysis of the block on the incline you need to use vector's components.

 

[sb_4000]

what are the vector components again?

 

[Pyrrhus]

Well, You need to study vectors and then try another attempt to solve the problem.

 

[sb_4000]

isnt the vector components like F_x = F*cos(theta) F_y = F*sin(theta)?

 

[Pyrrhus]

I'll leave the answer in vector equations. I've to go.

Incline Block:

$$\vec{N} + m_{2}\vec{g} + \vec{T} + \vec{F_{f}} = m_{2} \vec{a}$$

Hanging Block:

$$m_{1}\vec{g} + \vec{T} = m_{1} \vec{a}$$

Also Remember

$$F_{f} = \mu N$$

 

[Pyrrhus]

isnt the vector components like F_x = F*cos(theta) F_y = F*sin(theta)?

Yes they are, try to see how you can apply them, and see if you can understand the equations i left.

 

[sb_4000]

Thanks a lot..Ill try to figur it out.

 

[sb_4000]

Oene question, was the second part of my question correcty, it was 4N..the magnitude of tension..thanks again

 

[Pyrrhus]

sb4000, did you figure it out?

 

[sb_4000]

Hi,
Im still having problems solving part a, but I did part b, and I got T=39.19.

 

[Pyrrhus]

I will like to see your equations for both blocks.

 

[sb_4000]

T-m1gsin(theta) = m1a
-t+m2g=m2a

-(m1gsin(theta)+m1a)+m2g=m2a

the I got the acceleration a= m2g-m1gsin(theta)/m1+m2

T=m1gsin(theta)+m1a

this what I used to get T, my teacher had showed us a similar problem..

 

[Pyrrhus]

Your equation for the hanging block is correct, but in your equation for the incline block you forgot friction. Find tension using the equation from the hanging block.