2003 Leaving cert Vectors Question

ublinia2

Can somebody take a look at this vector question and explain how exactly they can get the answer F=mgsin(Theta) as per the marking scheme please.

The question states:

A person in a wheelchair is moving up a ramp at a constant speed.Their total weight is 900N.The ramp makes an angle of 10 degrees with the horizontal.
Calculate the force required to keep the wheelchair moving with a constant speed up the ramp.

Its on the examinations webpage at the link below.

http://www.examinations.ie/archive/exampapers/2003/LC021ALP1EV.pdf

Thanks

dudara

I can't draw a diagram at the moment, so i hope that I explain this ok in words.

Weight always acts vertically downwards, however, your x-direction is along the slope, which is at an angle of 10 deg.

You have to resolve the weight vector into components parallel to and perpendicular to the new x-y frame.

The weight vector makes an angle of 10 degrees with the y-axis or 80 with the x-axis. Therefore the component of the weight vector along the x-axis is WCos80 or WSin10. (Once you know the angle that a vector v makes with the horizontal, then the x-component is vCos(theta) anf the y-component is vSin9theta)).

If someone can supply a diagram, it will really help.

Sev

Ive already answered this question before actually on boards.
I gave a rather detailed response with diagram.. here's the link.

http://www.boards.ie/vbulletin/showthread.php?p=1419107

bazpaul

Nice work Sev,

a little long winded i might add