Question

It takes 1506 lb of force to pull a 5800 lb truck out of the snow. What is the coefficient of static friction?

Answer 1

Answer on Question #43326 – Physics - Mechanics | Kinematics | Dynamics

It takes 1506 lb of force to pull a 5800 lb truck out of the snow. What is the coefficient of static friction?

Solution

$$F = 1506 \text{ lb} \cdot \text{g} - \text{initial force};$$ $$m = 5800 \text{ lb} - \text{mass of the truck};$$

Second Newton's law along the X-axis:

$$F - F_{\text{frict}} = 0$$

Second Newton's law along the Y-axis:

$$mg = N$$

Formula for the friction force ($k$ – coefficient of static friction):

$$F_{\text{frict}} = N \cdot k = mg \cdot k$$ $$(2) \text{ in}(1):$$ $$F - mg \cdot k = 0$$ $$k = \frac{F}{mg} = \frac{1506 \text{ lb} \cdot \text{g}}{5800 \text{ lb} \cdot \text{g}} = \frac{1506}{5800} = 0.26$$

Answer: coefficient of static friction is equal to 0.26.

http://www.AssignmentExpert.com/