IIT JEE , AIEEE Forum - Physics , Mechanics

lonewolf07

Plz solve me this ????

Three forces are applied to a square plate as shown in the figure. Determine modulus, direction, and the point of application of resultant force, if this point is taken on side BC.

*f'' = f<root>2

Answer: F = 2f, parallel to diagonal AC, applied to midpoint of BC.

pink_ele

simply add three vectors

f(along bc) + f(along bd)=f

2(along ab)

f root2(along ab)+f root2(along ad)= 2f along ac

rite

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elessar_iitkgp

Net horizontal force, H = f + f

2cos 45 = 2f
Net vertical force, V = f - f

2 sin45 = 0
Hence the net force is horizontal, ie, parallel to AC.
Magnitude of resultant force, R = 2f
Let this resultant force be applied at a point x distant from C on the side BC.
Then the torque due to this force at any point must equal the sum of torques of the forces shown.
Taking torques about the point of intersection of the diagonals,
2f xsin45 = f(L/

2) + f(L/

2) - (f

2cos 45)((L/

2)
x = L/2
where L is the length of the side of the square.
Hence the resultant is applied at the midpoint of the side BC

iitkgp_bipin

Well done elessar_iitkgp.

Net horizontal component = f + (

2f)cos45⁰ = 2f
Net vertical component = f - (

2f)sin45⁰ = 0

Now apply the torque equation as done by elessar_iitkgp.
This question is simple and involves basics of vectors.

shwetadushyant

estimate the speed of vertically falling raindrops from the following data.

a. radius of drops =0.02 cm,viscocity of air=1.8*10^-4 poise,g=9.9,&density of water =1000kg/m^3.

elessar_iitkgp

Thanx bipin

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