Let's break down and solve the problem step by step: --- **Problem Restatement:** A box of mass \( m = 60 \) kg, height \( h = 4 \) m, width \( w = 3 \) m, is on a flat surface with a coefficient of static friction \( \mu_s = 0.4 \). A force \( F_{\text{push}} \) is applied horizontally at the top of the box. We increase \( F_{\text{push}} \) until the box either slides or tips. **Question:** Which happens first: sliding or tipping? **Justify your answer.** --- ### Step 1: Find the Force Needed to Slide the Box The maximum static friction force is: \[ F_{\text{friction, max}} = \mu_s N = \mu_s mg \] Plug in the numbers: - \( \mu_s = 0.4 \) - \( m = 60 \) kg - \( g = 9.8 \) m/s² \[ F_{\text{friction, max}} = 0.4 \times 60 \times 9.8 = 0.4 \times 588 = 235.2 \text{ N} \] So, the box will **slide** if \( F_{\text{push}} > 235.2 \) N. --- ### Step 2: Find the Force Needed to Tip the Box Tipping occurs when the torque due to \( F_{\text{push}} \) about the edge of the box equals the torque due to gravity. - **Axis of rotation**: bottom edge of the box. - **Torque due to \( F_{\text{push}} \)**: \( F_{\text{push}} \) acts at height \( h \). Perpendicular distance from pivot = \( h \). - **Torque due to gravity**: Weight acts at center of mass, which is at \( w/2 \) from the pivot. Set the torques equal at the tipping point: \[ F_{\text{push}} \cdot h = mg \cdot \frac{w}{2} \] Plug in numbers: - \( h = 4 \) m - \( m = 60 \) kg - \( g = 9.8 \) m/s² - \( w = 3 \) m \[ F_{\text{push}} \cdot 4 = 60 \times 9.8 \times \frac{3}{2} \] \[ F_{\text{push}} \cdot 4 = 60 \times 9.8 \times 1.5 = 60 \times 14.7 = 882 \] \[ F_{\text{push}} = \frac{882}{4} = 220.5 \text{ N} \] --- ### Step 3: Compare Forces for Sliding and Tipping - **Sliding:** \( F_{\text{push}} = 235.2 \) N - **Tipping:** \( F_{\text{push}} = 220.5 \) N **Tipping force is less than sliding force.** --- ### Step 4: Final Answer & Justification \[ \boxed{\text{The box will tip over before it slides.}} \] **Justification:** The force required to tip the box (220.5 N) is less than the force required to overcome static friction and slide the box (235.2 N). Therefore, as the force increases, the box will tip first.