The maximum amount of pulling force a truck can apply when driving on concrete is 9500 N . If the coefficient of static friction between a trailer and concrete is 0.8 , what is the heaviest that the trailer can be and still be pulled by the truck?

To determine the maximum weight of the trailer that can be pulled by the truck, we can use the relationship between the force of static friction and the normal force. The force of static friction (\(F_f\)) can be calculated using the formula: \[ F_f = \mu_s \cdot N \] where: - \(F_f\) is the force of static friction, - \(\mu_s\) is the coefficient of static friction (0.8 in this case), - \(N\) is the normal force, which is equal to the weight of the trailer (\(W_t\)) when on a flat surface. The weight of the trailer can be expressed as: \[ W_t = m_t \cdot g \] where: - \(m_t\) is the mass of the trailer, - \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)). Since the maximum pulling force the truck can apply is equal to the maximum static friction force, we can set up the equation: \[ F_{\text{max}} = F_f \] Substituting the expressions for \(F_f\) and \(W_t\): \[ 9500 \, \text{N} = \mu_s \cdot (m_t \cdot g) \] Now substituting \(\mu_s = 0.8\) and \(g = 9.81 \, \text{m/s}^2\): \[ 9500 \, \text{N} = 0.8 \cdot (m_t \cdot 9.81 \, \text{m/s}^2) \] Now, we can solve for \(m_t\): \[ 9500 = 0.8 \cdot 9.81 \cdot m_t \] Calculating \(0.8 \cdot 9.81\): \[ 0.8 \cdot 9.81 = 7.848 \] Now substituting back into the equation: \[ 9500 = 7.848 \cdot m_t \] Now, solving for \(m_t\): \[ m_t = \frac{9500}{7.848} \approx 1210.5 \, \text{kg} \] Thus, the heaviest that the trailer can be and still be pulled by the truck is approximately **1210.5 kg**.