Not solvable question civil engineering

They didn’t give me the radius of the wheel where external force F is acting .

How can I calculate the reactions at A,B .

When they didn’t give me the radius .

Is this question “ Wrong “ ?

Note the force F in the right down wheel

Can someone confirm that $$A_y = 0 , B_y = F$$

• Hard to tell what the question author had in mind. I would guess that they meant for the radius to be neglected. I. E. Assume it is zero. But that's just a guess. For anything definite you have to ask your instructor or whoever gave you the question Commented Nov 16, 2019 at 15:24
• I agree. If it's not given then assume (and state this in your answer) that the radius of the pulley is small in comparison with a and can be ignored. Commented Nov 16, 2019 at 15:26
• If it’s zero then the reaction $A_{y}=0$ ? Is it possible ? Commented Nov 16, 2019 at 15:34

Well I had to scratch my head on this one and dust off more than a few cobwebs. Now my understanding is that if you put a pulley on the ceiling and pass a rope through it to a 100 lbs weight and you hold that weight you are pulling on the free end of the rope with 100 lbs. If you are pulling straight down, it means you are are putting 100 lbs and the block is pulling down at 100 lbs and the pulley is exerting 200 lbs on the ceiling.

If you are pulling parallel to the ceiling then there is a 100 lbs down force and a 100 lbs horizontal force. or if you prefer a 141.4 lbs force at 45 degrees downward from the ceiling.

If my understanding is not valid, then the rest of this answer is not valid.

So I dropped this problem into S-Frame and placed the sum of the component horizontal and vertical loads as nodal loads. Because I can't just put 'a' in as a length I opted to make 'a' 10.

Now provided my understanding of pulleys is correct, then this model confirms there is no vertical load at A (point 1)

This is with the understanding that the vertical and horizontal components of th e 45 degree line break into roughly 70.7107 each

Sum of forces at point 7 is:

Fy=100 down from vertical + 70.7107 down from diagonal
Fx=70 right from diagonal


Sum of forces at point 4 is:

Fy=100 up from vertical


Sum of forces at point 5 is:

Fy= 100 down from vertical + 70.7107 up from diagonal
Fx= 70.7107 left from diagonal


We have the force $$(\sqrt 2/2)* F \$$ acting horizontally to the right at point H and horizontally to the left at point E , both horizontal components of the cord tension, F making a moment of $$\ M=(\sqrt 2/2)* F*a .$$

Now we equate sum of moments about A to zero and find reaction at B.

$$\Sigma M_A =0 \quad= \frac{(\sqrt 2/2)* F*a + F*2a}{2a} +R_B = 0$$

$$-R_B=F+ \sqrt 2/4* F$$

We ignored the radius of pulley as implied by absence of it in the diagram

Edit

After checking my answer I realized there was an arithmetic error. So I used the simple way of just summing the moments about A to simplify it.

We have two forces acting on the system: $$Fv \quad and \ F_{diagonal}$$

$$\Sigma M_A=0 \quad F* \frac{(3 \sqrt 2* a)}{2*2a}+ F* \frac{2a}{2a}+R_b=0$$ Note:$$(3 \sqrt2/2*a)= \text{distance from A to F(diag).}$$

$$-R_b= 3 *\sqrt 2/4*F+F$$

$$R_A=3 *\sqrt 2/4*F$$

Note: The sign reversal between the reactions indicates uplift reaction on A.

We ignored the radius of pulley as implied by absence of its dimension in the diagram. Please check my arithmetic.

• You could simplify the answer further to be Rb = -F(1+sqrt(2)/4) = -F(4+sqrt(2))/4. Commented Nov 17, 2019 at 4:14
• But why did you ignore the vertical components of the two pulleys Commented Nov 17, 2019 at 6:08
• Also should the reaction $R_B = F$ upwards since the only external force acting in the system is F downwards and $R_A = 0$ Commented Nov 17, 2019 at 6:10
• I tried your way taking into account the vertical components I get the same result $R_B = F$ Commented Nov 17, 2019 at 6:23
• @nick012000 are you sure your answer right ? Because I don’t think you calculated it right neglecting vertical forces. ~Nick012000 Commented Nov 17, 2019 at 13:00