Physics friction incline question

[Kevin1199959]

Homework Statement

Block B on a frictionless table top (1.5kg) is connected by a cord to block A (3.8kg) with friction on an incline of 40degrees above the horizontal.

Block B is being pulled at and angle 15degrees above the horizontal

Block A is moving up the incline at constant velocity.

Find Pulling force (on B at 15degrees), Tension in the cord, Friction and normal forces on A.

Homework Equations

F=ma
F=uN

The Attempt at a Solution

I found the X and Y components of block A, and i figured that Wx+Friction=F(on blobk B)

But i don't know how to find the rest.

TY in advance!

 

[tiny-tim]

Welcome to PF!

HI Kevin1199959! Welcome to PF!

(have a mu: µ and a degree: ° )

I found the X and Y components of block A, and i figured that Wx+Friction=F(on blobk B)

(i assume there's a pulley, to keep the cord level?)

You mean W_x+Friction = tension (in the cord)?

ok, so there are four forces on block B: tension, weight, normal reaction force, and pulling force …

draw a free body diagram and solve …

what do you get? ​

 

[Kevin1199959]

HI Kevin1199959! Welcome to PF!

(have a mu: µ and a degree: ° )(i assume there's a pulley, to keep the cord level?)

You mean W_x+Friction = tension (in the cord)?

ok, so there are four forces on block B: tension, weight, normal reaction force, and pulling force …

draw a free body diagram and solve …

what do you get? ​

Lol, Thanks ;)

Umm, as i mentioned, there is a force pulling on block b at 15° above horizontal, and if the system is moving at constant velocity, (a=0) the opposite forces have to be equal. Therefore left (in my case, friction+Wx of block A) = Right (pull at 15° on block B).

If I am right, Tension-Friction-Wx= 0 (ma=0) for block A and Force(15°)-Tension=0(ma=0) for block B

 

[tiny-tim]

Tension-Friction-Wx= 0 (ma=0) for block A and Force(15°)-Tension=0(ma=0) for block B

Yes (if you mean Force*cos15°) …

now put in the values of W_x and Friction and solve. ​

 

[Kevin1199959]

Yes (if you mean Force*cos15°) …

now put in the values of W_x and Friction and solve. ​

Yeah, i didn't mean *15°, i was just saying that it's the force at 15°. Sry :P

But i don't have the value of friction... How can i solve it?

 

[tiny-tim]

But i don't have the value of friction... How can i solve it?

Usual way …

find the normal force, and multiply by µ ​

 

[Kevin1199959]

Usual way …

find the normal force, and multiply by ​

But i don't have µ, I only have numbers that i gave in my first post...

 

[tiny-tim]

ooh, I never noticed that!

You'd better call the friction "µ", and get an answer with µ in it.

(or email the professor … I assume it's a misprint!)

 

[Kevin1199959]

But it was a question on the test lol... Here, i took a picture of the question in case i missed something.

Thx!

 

[tiny-tim]

hmm … I still think it's misprint!

 

[Kevin1199959]

hmm … I still think it's misprint!

It can't be... It was on the test and many people got it right