Recent content by Andy Salter

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    Dynamics: Forces in mechanism problem

    Ahh right, so only only if you had torques exerted on both links would the mass be sustained by both? Yeah I usually would have done those as dotted lines to show they are resultant forces, i drew that very quickly. Your pedantism has been very helpful /s
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    Dynamics: Forces in mechanism problem

    My FBD looked something like this:
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    Dynamics: Forces in mechanism problem

    Yes: $$\textbf{I'm just confused as to why there is no vertical component of the force at D}$$ Why's that? I would have thought it would be equally distributed? What about tangential acceleration? Why would this not be felt at both B and D? Cheers
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    Pressure at same height in same fluid?

    The fluid is static. ie. in equilibrium. Remember that you can think of pressure at a point as being exerted equally onto that point from all directions. So if the pressures were different at the bottom on the left and right, then there would be a net flow from high pressure to low pressure ie...
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    Pressure at same height in same fluid?

    If I'm not mistaken: Yes, it's because the tank is sealed. The water is at different heights so the pressure differential is given by: $$\rho_A g h_1 - \rho_w g h_2 - \rho_B g h_3 \\ = g(2\rho_A - 2\rho_w) \\ = 2g(\rho_A-\rho_w)$$ assuming ##\rho_A = \rho_B## its been a while since I've done...
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    Dynamics: Forces in mechanism problem

    Homework Statement Viewing a solution on the internet to this question, the free body diagram is drawn as: The forces here are then used to solve for ##F_D## and this is given as the answer to "the force which the bar exerts on the link at D". I'm just confused as to why there is no...
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    A question about friction coefficients....

    If a body is moving in one constant direction, is it considered stationary with respect to other directions? For example, consider a table. Let the surface of the table be defined by an x-y Cartesian plane. If the body is moving in the x direction on the table, and a force (< (μs * Fn)) is...
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    Vertical foce of fluid jet split in two

    Yes. I think I have the solution: ΣFy = ΣρQ(outlet) (because V1y = 0, inlet term of momentum equation is zero) for mass flow rate at outlet we only need to consider outlet 2 because velocity y-component of 3 is also zero. The mass flow rate at 2 is 2 kg/s and V2 = 1 so we have: ΣFy...
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    Vertical foce of fluid jet split in two

    Homework Statement A water jet hits a wall and splits into two streams (see attachment). The jet carries 3kg of water per second. The cross-sectional area of the flow at (2) is A = 2*10^-3 m^2. The average velocities at the outlets are V2 = V3 = 1m/s What is the net vertical force exerted by...
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    Find mass of load that falls off truck

    Ah yes, you are indeed right. Thankyou for your help :)
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    Find mass of load that falls off truck

    Fx = 19042 + Fgx = 19042 + FgSin(15) = 1.5m which gives the simultaneous equations: Fg = 9.81m Fg = (1.5m - 19042)/ Sin(15) With solutions: f = -46249 and m = 4714.52 Therefore: Δm = 7500 - 4714.52 = 2785.48 kg Yes?
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    Find mass of load that falls off truck

    Oh, wait, I see why that is wrong haha
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    Find mass of load that falls off truck

    My understanding was that Fx = Fgx = FgSin(15)
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    Find mass of load that falls off truck

    Fx = 19042 + Fgx = FgSin(15) Fy = 0 => FgSin(15) = Fx = F = ma = 1.5m => Fg = 1.5m/Sin(15)
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