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

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

(ii) Find μ

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

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

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

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

$\displaystyle Tcos30=16+51.96$

$\displaystyle T=67.96/cos30$

$\displaystyle T=78.5N$

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

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

(ii) Find μ

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

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

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

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

$\displaystyle Tcos30=16+51.96$

$\displaystyle T=67.96/cos30$

$\displaystyle T=78.5N$

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 $\displaystyle \mu$ ... should be 0.313

for part (b) ...

let $\displaystyle t$ = rope angle rel/horiz

$\displaystyle T$ = tension

$\displaystyle N$ = normal force

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

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

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

solve for $\displaystyle T$