Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread

When can you assume that the reaction forces at two supports are the same? Is it only when the questions tell you they are, otherwise always assume they are not the same magnitude?

only 6 marks?

here

if its is an isoceles triangle, and sinA=3/4, then drawing a triangle gives cosA as 3/4 as well i think?

cosa^2+sina^2 = 1.125 but i thought that was supposed to be equal to one

$\sin(a)=\frac{3}{4}$ and $\cos^{2}(a)=1-\sin^{2}(a)=\frac{7}{16}$ so $\cos(a)=\frac{\sqrt{7}}{4}$

yh i forgot that was just for right angle triangles

My favourite M1 problem ever!!! And home made of course.

Do yoy have the answer for a? Not sure ive done it right

Do yoy have the answer for a? Not sure ive done it right[/QUOTE

Answer in spoiler
Please say if anything wrong.

Do yoy have the answer for a? Not sure ive done it right[/QUOTE

Answer in spoiler
Please say if anything wrong.

oh i thought that as it says the rod remains in equilbrium, then the blocks remained in equilbrium, which isnt true

The force that the tension exerts on the pulley keeps the rod in equilibrium as it produces a moment.

so the resultant force of the tension acting on the rod is equal to the weight of the particle on the rod?

Can someone help me urgently please? <3

LOOK IN DIAGRAM!


So in this question where it says 2.5 -x

I labelled that part as x and the other part where it says x as 2.5 - x, so essentially I switched them around

I did the calculation and got x = 2 MY WAY, but the mark scheme way (not switched around) got x = 0.5

Am I not aloud to label the end part as 2.5 - x or something?

No yours is right, you just have to mention that it is 2 metres from M to B or something like that.

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In June 2014 r , why is it in part B the S is doubled