# Mechanics 1 question. Forces/friction/Newtons 2nd Law

• Jan 8th 2010, 03:09 AM
NathanBUK
Mechanics 1 question. Forces/friction/Newtons 2nd Law
Hi, just another question which im slighly unsure about but Ill see if you can help me.

A block of mass 20kg is being pulled along a rough horizonal surface by a rope inclined at an angle of 30degrees to the horizonal.

The coefficient of friction between the block and surface is μ. Model the block as a particle which slides on the surface.

(a) If the tension in the rope is 60 newtons, the block moves at a constant speed.

(i) Show that the magnitude of the normal reaction force acting on the block is 166N

I did $R+60cos60=20*9.8$ $R=196-60cos60$ $R=166N$ so I assume I got this bit correct.

(ii) Find μ

I did $F=60cos30$ $F=51.96$

F=μR 51.96=μx166 μ= $51.96/166$ μ=0.131

(b) If the rope remains at the same angle and the block accelerates at $0.8ms-2$, find the tension in the rope.

I said 'By Newtons 2nd Law Resultant Force = ma
so
$Tcos30-51.96=20*0.8$

$Tcos30=16+51.96$

$T=67.96/cos30$

$T=78.5N$

The last question im pretty sure I have no idea what Im talking about, and please excuse the maths talk, werent sure how to get all the symbols in correctly.

Thanks :)
• Jan 8th 2010, 04:51 PM
skeeter
Quote:

Originally Posted by NathanBUK
Hi, just another question which im slighly unsure about but Ill see if you can help me.

A block of mass 20kg is being pulled along a rough horizonal surface by a rope inclined at an angle of 30degrees to the horizonal.

The coefficient of friction between the block and surface is μ. Model the block as a particle which slides on the surface.

(a) If the tension in the rope is 60 newtons, the block moves at a constant speed.

(i) Show that the magnitude of the normal reaction force acting on the block is 166N

I did $R+60cos60=20*9.8$ $R=196-60cos60$ $R=166N$ so I assume I got this bit correct.

(ii) Find μ

I did $F=60cos30$ $F=51.96$

F=μR 51.96=μx166 μ= $51.96/166$ μ=0.131

(b) If the rope remains at the same angle and the block accelerates at $0.8ms-2$, find the tension in the rope.

I said 'By Newtons 2nd Law Resultant Force = ma
so
$Tcos30-51.96=20*0.8$

$Tcos30=16+51.96$

$T=67.96/cos30$

$T=78.5N$

The last question im pretty sure I have no idea what Im talking about, and please excuse the maths talk, werent sure how to get all the symbols in correctly.

Thanks :)

(a) (ii) your equation is correct, but check your calculation for $\mu$ ... should be 0.313

for part (b) ...

let $t$ = rope angle rel/horiz

$T$ = tension

$N$ = normal force

$T\cos{t} - \mu N = ma$

$N = mg-T\sin{t}$

$T\cos{t}-\mu(mg-T\sin{t}) = ma$

solve for $T$