A particle of mass 1.5kg rests on a rough plane inclined at 45° to the horizontal. It is maintained in equilibrium by a horizontal force of p newtons. Given that the coefficient of friction between the particle and the plane is (1/4), calculate the value of p when the particle is on the point of moving i. down the plane ii. up the plane [take g=10ms^(−2) ].


Question and Answers Forum

All Questions Topic List
Vector Questions
Previous in All Question Next in All Question
Previous in Vector Next in Vector
Question Number 54578 by pieroo last updated on 07/Feb/19

$A particle of mass 1 . 5 kg rests on a rough$ $plane inclined at 45 ° to the horizontal .$ $It is maintained in equilibrium by a$ $horizontal force of p newtons . Given$ $that the coefficient of friction between$ $the particle and the plane is \frac{1}{4}, calculate$ $the value of p when the particle is on$ $the point of moving$ $i . down the plane$ $ii . up the plane$ $[take g = {10 ms}^{- 2}] .$
Commented by pieroo last updated on 07/Feb/19

$Please i need your help .$
	Answered by mr W last updated on 07/Feb/19

	$N = m g \cos α + p \sin α$ $(i)$ $m g \sin α - p \cos α = μ N = μ (m g \cos α + p \sin α)$ $\Rightarrow p = \frac{m g (\sin α - μ \cos α)}{μ \sin α + \cos α}$ $= \frac{1.5 \times 10 (1 - \frac{1}{4})}{\frac{1}{4} + 1} = 9 N$ $(i i)$ $p \cos α - m g \sin α = μ N = μ (m g \cos α + p \sin α)$ $\Rightarrow p = \frac{m g (\sin α + μ \cos α)}{\cos α - μ \sin α}$ $= \frac{1.5 \times 10 (1 + \frac{1}{4})}{1 - \frac{1}{4}} = 25 N$
	Commented by mr W last updated on 07/Feb/19

	Commented by mr W last updated on 07/Feb/19

	$f i g u r e a b o v e s h o w s c a s e (i) .$
	Commented by pieroo last updated on 08/Feb/19

	$thanks sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum