ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ahmed on April 20, 2009, 04:17:32 pm

Title: Need h/w help asap
Post by: ahmed on April 20, 2009, 04:17:32 pm: A force p, parallel to a surface inclined 15 deg. above the horizontal, acts on a 45-N block, as shown below in the attached pic.
The coefficients of friction for the block and surface are static u_s=0.5 and u_k=0.34. If the block is initially at rest,
determine the magnitude and direction of the frictional force acting on the block for magnitudes of p of
a) 5N
b)8N
c)15N
Title: Re: Need h/w help asap
Post by: Mao on April 20, 2009, 07:45:13 pm: the block weighs 45 newtons. At an angle of inclination of 15 degrees, it has $45\sin 15^o$ N parallel to the surface, and $45\sin 15^o$ N perpendicular to the surface. To overcome static friction, the sum of forces in the direction of motion must be greater than maximum static friction. Since P is acting downhill, static friction must act uphill.

Since P is acting in the horizontal (i) direction, it has a component parallel and perpendicular to the plane as well.

Resolving forces perpendicular to the plane, P and the normal reaction force N pulls the block away from surface, gravity pulls down:

$N + P\sin 15^o = 45\cos 15^o\implies N = 45 \cos 15^o - P\sin 15^o$

Resolving forces parallel to the plane

If the total down-slope forces are greater than static friction: $P\cos 15^o + 45\sin 15^o > \mu_s \cdot N \implies P \cos 15^o + 45\sin 15^o > 0.5\cdot ( 45 \cos 15^o - P\sin 15^o )$
Then $F_{net} = P\cos 15^o + 45\sin 15^o - \mu_k N$ , where resistive forces $F_{k} = \mu_k N$

Otherwise, $F_s = P\cos 15^o + 45\sin 15^o$ (total resistive force is equal to total downslope force, no net motion)

sub in the values of P to see which of the above two scenario it fits.