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Math  /  Algebra

QuestionProblem solving process:
1. Draw a complete force diagram for each object. Include components and tick marks where appropriate.
2. Obtain the ΣFx\Sigma F_{x} and Σy\Sigma_{y} equations.
3. Substitute in formulas or known values.
4. Solve for unknown.
1. You are trying to push a 100 kg crate across a level floor. The coefficient of static friction between the crate and the floor is 0.25 , and the kinetic coefficient is 0.20 . a. Draw a force diagram under the figure. ΣFy=Fr−Fg=ma=0FN=Fg=9.80100=980\begin{array}{l} \begin{aligned} \Sigma F_{y}=F_{r}-F_{g} & =m a=0 \\ F_{N}=F_{g} & =9.80100 \\ & =980 \end{aligned} \end{array} b. What horizontal force must you apply to the crate to get the crate moving? c. What horizontal force must you apply to the crate to keep the crate moving with constant velocity? Kine tic C. FFK=0.20⋅980F_{F_{K}}=0.20 \cdot 980

Studdy Solution

STEP 1

1. The crate is initially at rest and requires a force to overcome static friction to start moving.
2. Once the crate is moving, kinetic friction applies, and a constant force is needed to maintain constant velocity.
3. The gravitational acceleration g g is approximately 9.8 m/s2 9.8 \, \text{m/s}^2 .

STEP 2

1. Draw a complete force diagram.
2. Calculate the normal force FN F_N .
3. Determine the force needed to overcome static friction.
4. Determine the force needed to maintain constant velocity against kinetic friction.

STEP 3

Draw a complete force diagram for the crate. Include: - Gravitational force Fg F_g acting downward. - Normal force FN F_N acting upward. - Applied force Fa F_a acting horizontally. - Frictional force Ff F_f acting opposite to the direction of Fa F_a .

STEP 4

Calculate the normal force FN F_N . Since the crate is on a level floor and there is no vertical acceleration, the normal force equals the gravitational force:
FN=Fg=m⋅g=100 kg×9.8 m/s2=980 N F_N = F_g = m \cdot g = 100 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 980 \, \text{N}

STEP 5

Determine the force needed to overcome static friction. Use the formula for static friction:
Ffs=μs⋅FN F_{f_s} = \mu_s \cdot F_N
Where μs=0.25 \mu_s = 0.25 is the coefficient of static friction.
Ffs=0.25×980 N=245 N F_{f_s} = 0.25 \times 980 \, \text{N} = 245 \, \text{N}
This is the minimum horizontal force needed to start moving the crate.

STEP 6

Determine the force needed to maintain constant velocity against kinetic friction. Use the formula for kinetic friction:
Ffk=μk⋅FN F_{f_k} = \mu_k \cdot F_N
Where μk=0.20 \mu_k = 0.20 is the coefficient of kinetic friction.
Ffk=0.20×980 N=196 N F_{f_k} = 0.20 \times 980 \, \text{N} = 196 \, \text{N}
This is the horizontal force needed to keep the crate moving with constant velocity.
The horizontal force required to start moving the crate is 245 N 245 \, \text{N} , and the force required to keep it moving at a constant velocity is 196 N 196 \, \text{N} .

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