Finding Kinetic Friction

The normal force exerted on a pram is \(\text{100}\) \(\text{N}\). The pram's brakes are locked so that the wheels cannot turn. The owner tries to push the pram but it doesn't move. The owner pushes harder and harder until it suddenly starts to move when the applied force is three quarters of the normal force. After that the owner is able to keep it moving with a force that is half of the force at which it started moving. What is the magnitude of the applied force at which it starts moving and what are the coefficients of static and kinetic friction?

Step 1: Maximum static friction

The owner of the pram increases the force he is applying until suddenly the pram starts to move. This will be equal to the maximum static friction which we know is given by: \[f_s^{max} = \mu_sN\]

We are given that the magnitude of the applied force is three quarters of the normal force magnitude, so: \begin{align*} f_s^{max} &= \frac{\text{3}}{\text{4}}N \\ &= \frac{\text{3}}{\text{4}}(\text{100}) \\ &=\text{75}\text{ N} \end{align*}

Step 2: Coefficient of static friction

We now know both the maximum magnitude of static friction and the magnitude of the normal force so we can find the coefficient of static friction: \begin{align*} f_s^{max} &= \mu_sN \\ \text{75} &= \mu_s (\text{100}) \\ \mu_s &=\text{0,75} \end{align*}

Step 3: Coefficient of kinetic friction

The magnitude of the force required to keep the pram moving is half of the magnitude of the force required to get it to start moving so we can determine it from: \begin{align*} f_k &= \frac{\text{1}}{\text{2}}f_s^{max} \\ &= \frac{\text{1}}{\text{2}}(\text{75}) \\ & = \text{37,5}\text{ N} \end{align*}

We know the relationship between the magnitude of the kinetic friction, magnitude of the normal force and coefficient of kinetic friction. We can use it to solve for the coefficient of kinetic friction: \begin{align*} f_k &= \mu_kN \\ \text{37,5} &= \mu_k (\text{100}) \\ \mu_k &=\text{0,375} \end{align*}