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I would appreciate greatly your assistance on the following problem:

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The answer given in the book is:

Acceleration of wedge: $0.7 \frac{m}{s^2}$.
Reaction between wedge and horizontal plane: $173.25 N$.

I would like to understand how these values are derived. So far I have attempted to resolve the forces in the system as follows:

Eqn 1) $R_{5} - R_{3} = 10\sqrt{2}a$ [wedge (i)]
Eqn 2) $R_{5} - T = 5b - 5\sqrt{2}a$ [M5 (i)]
Eqn 3) $5\sqrt{2}g - R_{5} - T = 5b$ [M5 (j)]
Eqn 4) $T - R_{3} = 3b-3\sqrt{2}a$ [M3 (i)]
Eqn 5) $T + R_{3} - 3\sqrt{2}g = 3b$ [M3 (j)]

Any help at all would be very much welcome.

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  • 1
    You haven’t included the string tension among the forces acting on the wedge. The two tensions cancel in the horizontal direction, so you can still come up with the right value for the wedge’s acceleration, but there’s no hope of finding the reaction force without taking the string tension into account. – 2017-01-25
  • 0
    Thank you very much for the insightful comment. I've tried to calculate T as follows: $T - \frac{3}{\sqrt{2}}g = 3b$, $\frac{5}{\sqrt{2}}g - T = 5b$, $b = \frac{1}{4\sqrt{2}}g = 1.732$, $T = 25.986$. I believe that the force at the pulley should therefore be ~36.75N. I'm still no closer to the magic number of 173.25N though. – 2017-01-25
  • 1
    Sounds like you’re still guessing. The value of $T$ isn’t really important yet. What matters more is how $T$ fits into Eq 1 and the equation that you haven’t written down yet for the vertical components of the forces acting on the wedge. Once you have those two equations right, the rest is straightforward algebra—no guessing or trial and error (as you stated in an earlier edit) required. – 2017-01-26
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    That's just the explanation I needed - thank you! Vertical components of forces on wedge are: $1\sqrt{2}R_{5} + 1\sqrt{2}R_{3} + 10g + 2\sqrt{2}T$; solve for $R_{5}$, $R_{3}$, and $T$ then substitute and the required number emerges at last. I'm very grateful indeed for your help :) – 2017-01-27
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    Great! Remember that there’s a “hidden” $T$ term in Eq 1. The horizontal components cancel in this case because of the equal angles, but they’re there all the same. – 2017-01-27

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