Question

A 45kg box is pulled across a horizontal floor. A rope attached to the box makes an angle of 30 degrees with the horizontal and the coefficient of friction between the box and the floor is 0.7. What is the tension in the rope in order to pull the box at a constant rate?

Answer 1

Question

We will have the horizontal forces (resultant force equals to the zero because the box is pulling at a constant rate $\Rightarrow a = 0$):

$$\begin{array}{l} N = m \cdot g \\ F_{\text{friction}} = \mu \cdot N = \mu \cdot m \cdot g = 0.7 \cdot 45 \cdot 9.8 = 308.7 \, N \\ T_{\text{horizontal}} = T \cdot \cos 30{}^\circ \\ T_{\text{horizontal}} - F_{\text{friction}} = m \cdot a = 0 \Rightarrow T_{\text{horizontal}} = F_{\text{friction}} \Rightarrow \\ \Rightarrow T \cdot \cos 30{}^\circ = F_{\text{friction}} \Rightarrow T = \frac{F_{\text{friction}}}{\cos 30{}^\circ} = \frac{308.7}{\cos 30{}^\circ} \approx 356.5 \, N \\ \end{array}$$

Answer: 356.5 N.