Force Problem Question

Question

A person holds a 100 kg. cart to keep it from sliding down the 30 degree inclined plane. Compute for the force exerted by the person if the coefficient of kinetic friction is 0.02.

First convert the 100 kg to newtons which is now 980 N so that is the force weight. then what do i do next?

Answer 1

Nice and easy and to the point:

Work in an axis system parallel and perpendicular to the plane...

This is a Static force balance problem.

Force down plane = -force up plane   plug in the symbols...

m*g*sin(30) = F(pull) + mu*m*g*cos(30)    plug in the numbers

980 * (1/2) - 980 * Sqrt(3)/2 * .02 =

F(pull) = 473 N

-Fred

Always check my aritmatic, but the method is A OK.  Actually A+ OK  :)

Answer 2

the accelaration of the body on inclined plane is :

a=gX sin(30degree) - f/m

therefore the force with which the body is sliding down is :

ma =m X  g X sin(30degree) - f/m x m

=ma = mgsin(30degre) - f

where,

f = friction force

m= mass of the cart= 100kg

g= 9.8m/s^2

a= accelaration of the cart

according to newtons law,

F= ma

therefore,

F= mgsin(30degee) - f--------------------(1)

the friction can be taken ot as:

mew = friction / normal reaction

0.02 = f / 980

f = 0.02 X 980

f= 19.6N

equation (1)becomes

F = 100 x 9.8 (0.5) - 19.6

F = 29.4N is the force with which the cart is sliding down

the same force must be applied by the boy to keep the cart at rest