Static friction of blocks on a wire

[MisterOL]

Homework Statement

2 blocks are connected with wire each on its own angle area. Body 1 (B1) has weight 900N, angle 30, µ = 0,35
B2 wight = 1100N, angle 45 and µ=0.30
There is no friction on at the T ( picture )
How big F has to be to pull block1 down following shown path?

Homework Equations

SUM FY=0
SUM Fx=0
Ff = µ * N

The Attempt at a Solution

I separated system and tok a look at block 2
- since there is 45 degrees angle between G2x and G2y they have to be same in size ( 1110 / 2 = 550 N )
- Since N2 = G2y, N2 is 550N as well. (Sum Fy=0)
- Since the point is to glide block nr 1 down that means that bloock 2 has to go up hence Friction force (F2) poiniting down or along the 45 degres plate
- If Ff = µ * N that means Ff2 = 0.30 x 550 which is 165
- Since G2x and FF2 pointing in the same direction (Sum Fx=0) there has to be an F force that pulls block 2 towards T and it is F2 = Ff2 + G2x which is 165 + 550 = 715 N

Now back to block 1
G1x = G * sin30 = 900 * sin30 = 450 N
G1y = G * cos30 = 779,42 N
N1 = 779, 42 N
Ff1 = 0,35 * 779,42 = 272,78 N
Sum Fx = 0 gives
+ Ff1 - F - G1x = 0 => F= Ff1-G1x => 272,78 - 450 = -177,22
This also means that F2 force from block 2 makes somehow impact.

The correct answer is F = 834 N and I have tried different aproaches but no luck .
Can someone that understand this perticular part please take a look at this problem ?

Help appritiated :)

 

[PhanthomJay]

Homework Statement

2 blocks are connected with wire each on its own angle area. Body 1 (B1) has weight 900N, angle 30, µ = 0,35
B2 wight = 1100N, angle 45 and µ=0.30
There is no friction on at the T ( picture )
How big F has to be to pull block1 down following shown path?

Homework Equations

SUM FY=0
SUM Fx=0
Ff = µ * N

The Attempt at a Solution

I separated system and tok a look at block 2
- since there is 45 degrees angle between G2x and G2y they have to be same in size ( 1110 / 2 = 550 N )

they have to be the same , but they are not 550 N each. Don't forget that the x comp of the gravity force is mgsintheta

- Since N2 = G2y, N2 is 550N as well. (Sum Fy=0)

same error

- Since the point is to glide block nr 1 down that means that bloock 2 has to go up hence Friction force (F2) poiniting down or along the 45 degres plate
- If Ff = µ * N that means Ff2 = 0.30 x 550 which is 165

make correction for N

- Since G2x and FF2 pointing in the same direction (Sum Fx=0) there has to be an F force that pulls block 2 towards T and it is F2 = Ff2 + G2x which is 165 + 550 = 715 N

the 'F' force is the _______in the rope

'
Now back to block 1
G1x = G * sin30 = 900 * sin30 = 450 N
G1y = G * cos30 = 779,42 N
N1 = 779, 42 N
Ff1 = 0,35 * 779,42 = 272,78 N
Sum Fx = 0 gives
+ Ff1 - F - G1x = 0 => F= Ff1-G1x => 272,78 - 450 = -177,22

you left out one of the forces acting along the plane...which one?

This also means that F2 force from block 2 makes somehow impact.

The correct answer is F = 834 N and I have tried different aproaches but no luck .
Can someone that understand this perticular part please take a look at this problem ?

Help appritiated :)

Draw good free body diagrams, identify all forces acting on the block. You'll have to solve 2 equations with 2 unknowns.

 

[MisterOL]

Hi Jay thanks for help :9
Ok here we go again
1) G2x = G * sin45 = 1100 * 0,707 = 777,82 N => G2x = G2y = N2
2) Ff2 = 0.30 * 777,82 = 233, 35 N
3) SUM Fx=0 for B1 => +Ff2 +G2x - F2 => F2 = 777,82 + 233.35 = 1011,17 N
4) G1x = G * sin30 = 450 N
5) G1y = G* cos30 = 779,42 N => G1y = N1
3) Ff1 = 0,35 * 779,42 = 272,78 N
6) F1 = Ff1 - G1x + F2 => 272,78 -450 + 1011,17 = 833.95 apx 834 :)

 

[PhanthomJay]

Nice work! Just be sure that you understand that F2 is the tension in the rope connecting thy 2 blocks. I'd call it T1=T2 =T =1011 N.