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# Question about the equilibrium of a rigid body?

Screenshot of question & diagram - https://prnt.sc/tgt9dm

I am not sure how to get force P.

I tried to use ΣM = 0 at point A: -(750 x 1) - (750cos36.9 x 2.5) = (P x 0.8)

However, my answer for P is 2811 N and the correct answer is 982 N.

Any help is appreciated.

### 2 Answers

- FiremanLv 71 month agoFavorite Answer
Let the components of R at A is Rx & Ry respectively in horizontal & vertical directions, Given:

At D: Px = 12P/13 & Py = 5P/13

At C: Fy = 4/5 x 750 = 600 N & Fx = 3/5 x 750 = 450 N

By balancing the the forces vertically & Horizontally:

=>Ry + 5P/13 = 75g + 600

=>Ry + 5P/13 = 1335 --------------(i)

& Rx = 450 + 12P/13 ---------------(ii)

By taking torque at A:

=>735 x 1 + 600 x 2.5 = 12P/13 x 0.8 + 5P/13 x 4

=>29.6P/13 = 2235

=>P = 981.59 ≈ 982 N

Thus By (i):

=>Ry = (5 x 982)/13 = 1335

=>Ry = 957.31 N

& By (ii):

=>Rx = 450 + (12 x 982)/13 = 1356.46 N

Thus By R = √[Rx^2 + Ry^2]

=>R = √[(1356.46)^2 + (957.31)^2]

=>R = 1660.25 ≈ 1660 N

& By tanθ = Ry/Rx = 957.31/1356.46 = tan35.2*

=>θ = 35.2*

- NCSLv 71 month ago
First, notice that the triangle associated with the load at C is a 3-4-5 triangle, and the one associated with the pulley is a 5-12-13 triangle.

For rotational equilibrium of the apparatus, we need to have a zero moment about A (well, about all points actually):

ΣMa = 0 = P*[(12/13)*0.8m + (5/13)*4.0m] -750N*(4/5)*2.5m - 75kg*9.8m/s²*1m

solves to

P = 982 N ◄

For horizontal equilibrium, the force at A must be such that

Fax = (12/13)*982N + 750N*(3/5) = 1356 N

and for vertical equilibrium

Fay = -(5/13)*982N + 750N*(4/5) + 75kg*9.8m/s² = 957 N

giving the reaction force at A a magnitude of

|Fa| = √(1356² + 957²) N = 1660 N ◄

and a direction of

Θ = arctan(Fay / Fax) = 35.2º ◄

above "east"