Determining the force of friction can be utilizing the force of friction equation or figuring out the component of the weight force that must be opposed to prevent motion. Use these practice problems below to practice finding the force of friction in different scenarios.
| Problem | Picture | Given/Find | Equation | Solution |
|---|---|---|---|---|
| A 51.0 kg desk moves at constant velocity across a hardwood floor (µk = 0.200). What is the force of kinetic friction on the desk? |
|
\(\small\mathsf{ m = 51.0 kg }\) \(\small\mathsf{ \mu_k = 0.200 }\) |
\(\small\mathsf{ \Sigma F = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} = 0 }\) \(\small\mathsf{ F_{friction} = \mu F_{normal} }\) |
\(\small\mathsf{ \overrightarrow{F}_{normal} = -\overrightarrow{F}_{weight} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = -(51.0 \text{ kg})(-9.81 \text{ m/s}^2) }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = 500.31 \text{ N} }\) \(\small\mathsf{ F_{friction} = (0.200)(500.31 \text{ N}) = 100. \text{ N} }\) |
| A block of wood whose mass is 5.00 kg rests on a horizontal floor. If the coefficient of kinetic friction is 0.520, what is the force of kinetic friction? |
|
\(\small\mathsf{ m = 5.00 kg }\) \(\small\mathsf{ \mu_k = 0.520 }\) |
\(\small\mathsf{ \Sigma F = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} = 0 }\) \(\small\mathsf{ F_{friction} = \mu F_{normal} }\) |
\(\small\mathsf{ \overrightarrow{F}_{normal} = -\overrightarrow{F}_{weight} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = -(5.00 \text{ kg})(-9.81 \text{ m/s}^2) }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = 49.05 \text{ N} }\) \(\small\mathsf{F_{friction} = (0.520)(49.05 \text{ N}) = 25.5 \text{ N} }\) |
| A 242 kg statue initially at rest on a horizontal floor requires a 2044 N horizontal force to set it in motion. What is the coefficient of static friction? |
|
\(\small\mathsf{ m = 242 kg }\) \(\small\mathsf{ \overrightarrow{F}_{applied} = 2044 \text{ N} }\) |
\(\small\mathsf{ \overrightarrow{F}_{applied} = -\overrightarrow{F}_{friction} }\) \(\small\mathsf{ \Sigma F = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} = 0 }\) \(\small\mathsf{ \mu = \frac{F_{normal}}{F_{friction}} }\) |
\(\small\mathsf{ \overrightarrow{F}_{friction} = -2044 \text{ N} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = -(242 \text{ kg})(-9.81 \text{ m/s}^2) = 2374 \text{ N}}\) \(\small\mathsf{ \mu_s = \frac{2044 \text{ N}}{2374 \text{ N}} = 0.861}\) |
| A 242 kg statue on a horizontal floor requires a 1766 N force that keeps it moving at a constant velocity. What is the coefficient of kinetic friction? |
|
\(\small\mathsf{ m = 242 kg }\) \(\small\mathsf{ \overrightarrow{F}_{applied} = 1766 \text{ N} }\) |
\(\small\mathsf{ \overrightarrow{F}_{applied} = -\overrightarrow{F}_{friction} }\) \(\small\mathsf{ \Sigma F = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} = 0 }\) \(\small\mathsf{ \mu = \frac{F_{normal}}{F_{friction}} }\) |
\(\small\mathsf{ \overrightarrow{F}_{friction} = -1766 \text{ N} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = -(242 \text{ kg})(-9.81 \text{ m/s}^2) = 2374 \text{ N}}\) \(\small\mathsf{ \mu_s = \frac{1766 \text{ N}}{2374 \text{ N}} = 0.744}\) |
