Question

A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration the horse can give if the coefficient of static friction between the horse’s hooves and the road is 0.894? Answer in units of m/s 2 .

Answer 1

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

$F_f = \mu N$

Where,

$\mu =$ Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

$m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894$

By condition of Balance the friction force must be equal to the total net force, that is to say

$F_{net} = F_f$

$m_{total}a = \mu m_hg$

$(m_h+m_c)a = \mu*m_h*g$

Re-arrange to find acceleration,

$a= \frac{\mu*m_h*g}{(m_h+m_c)}$

$a = \frac{0.894*242*9.8}{(242+224)}$

$a = 4.54 m/s^2$

Therefore the acceleration the horse can give is $4.54m/s^2$