Author Topic: m1 relly fast question  (Read 865 times)

Offline 7ooD

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m1 relly fast question
« on: March 24, 2010, 12:16:29 am »
ok i got the question and the solution but just a part i dnt understand

the question says :

a body of mass 2kd is held in the limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a horizontal force x. the coefficient of friction between the body and the plane is 0.2. modeling the body as a particle find x when the body is on the point of slipping a)up the plane b) down the plane

for this part i attached the solution or the way to solve it

the one i cnt solve is

the force in the previous question is now replaced by a force y at an angle of 45 degrees to the horizontal. find y when the body is on the point of slipping

a) up the plane b) down the plane

the answers should be a)43N b)3.67N
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Offline Saladin

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Re: m1 relly fast question
« Reply #1 on: March 24, 2010, 04:46:36 am »
Is this question from the Edexcel M1 book, and if so, please tell me the chapter number and question number.

Offline astarmathsandphysics

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Re: m1 relly fast question
« Reply #2 on: March 24, 2010, 07:30:08 am »
There is more than one way for a force to act at 45 degrees - it could act above the horizontal or below the horizontal. Which one is this?

Offline 7ooD

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Re: m1 relly fast question
« Reply #3 on: March 24, 2010, 10:29:24 am »
chp 4 ex 4e q 5 and 6 astar thats wat is written in the question :d
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Offline astarmathsandphysics

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Re: m1 relly fast question
« Reply #4 on: March 24, 2010, 02:40:13 pm »
res perp to plane
R-2gcos25-Ycos25=0 (1)
parallel to plane Y sin25-2gsin20-0.2R=0 (1)+5*(2)
Y(5sin25-cos25)-(2gcos25=10gsin25)=0
Y=42.49

cant scan a diagram now sorry.

res perp to plane
res perp to plane
R-2gcos25-Ycos25=0 (1)
parallel to plane Y sin25-2gsin20+0.2R=0 (1)-5*(2)
Y(-5sin25-cos25)+(10gsin20-2gcos25)=0
Y=-5.1 ie up