Question

A friction force of 280 N exists between a cart and the path. If the force of reaction is 766 N, what is the minimum action force needed to set the cart in motion?

Answer 1

force of friction on the cart is given as

$F_f = 280 N$

here we also know the reaction force due to surface

$R = 766 N$

so we can say reaction force is given as

$R = \sqrt{F_n^2 + F_f^2}$

$766 = \sqrt{F_n^2 + 280^2}$

$F_n^2 = 766^2 - 280^2$

$F_n = 713 N$

now by force balance we will say

$F_y + F_n = mg$

$F_y = mg - F_n$

$F_x = F_f$

also we know that

$F_f = \mu * F_n$

$280 = \mu * 713$

$\mu = 0.39$

now minimum force required to set this into motion

$F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$

here we know that

$mg = F_n = 713 N$

$F_{min} = \frac{0.39* 713}{\sqrt{1 + 0.39^2}}$

$F_{min} = 259 N$

So it will require 259 N minimum force to move it