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aev [14]
3 years ago
7

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 forc

e needed to set the cart in motion?
Physics
1 answer:
MatroZZZ [7]3 years ago
3 0

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

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