# Static friction of blocks on a wire

1. Jan 31, 2010

### MisterOL

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
SUM FY=0
SUM Fx=0
Ff = µ * N

3. 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 :)

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2. Jan 31, 2010

### PhanthomJay

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
same error
make correction for N
the 'F' force is the _______in the rope
you left out one of the forces acting along the plane.....which one?
Draw good free body diagrams, identify all forces acting on the block. You'll have to solve 2 equations with 2 unknowns.

3. Jan 31, 2010

### 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 :)

4. Jan 31, 2010

### 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.