An object with a mass of #12 kg# is lying still on a surface and is compressing a horizontal spring by #1/3 m#. If the spring's constant is #6 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?
We can use Newton's second law to solve.
Diagram:
We have the following information:
#|->m=12" kg"# #|->Deltas=-1//3" m"# #|->k=6" kg"//"s"^2# #|->g=9.81"m"//"s"^2#
We can sum the forces parallel and perpendicular:
#sumF_x=F_s-f_s=0#
#sumF_y=n-F_g=0#
- Note that the net force in both directions is 0 as the object remains at rest. The system is in a state of static equilibrium.
We also know:
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#f_(s"max")=mu_sn=>f_s<=mu_sn# -
#F_s=-kDeltas# -
#F_g=mg# Substituting the above equations into the statements we formed above:
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#n=mg# -
#-kDeltas-mu_(s"max ")mg=0# We can solve for
#mu_(s"max")# :#color(darkblue)(mu_(s"max ")=(-kDeltas)/(mg))# Submitting in our known values:
#mu_(s"max ")=-((6" kg"//"s"^2)(-1//3" m"))/((12" kg")(9.81"m"//"s"^2)# #=>~~0.017# Therefore,
#color(darkblue)(mu_s<=0.017)# .
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To find the minimum value of the surface's coefficient of static friction that keeps the 12 kg object still, we need to consider the forces acting on the object due to the compression of the spring. When the spring is compressed, it exerts a force on the object, which can be calculated using Hooke's Law:
[ F_{spring} = k \times x ]
where ( k ) is the spring constant (6 kg/s^2) and ( x ) is the compression distance (1/3 m). Substituting the values:
[ F_{spring} = 6 \times \frac{1}{3} = 2 , \text{N} ]
This force is what might cause the object to slide. For the object to remain stationary, the static friction force between the object and the surface must be equal to or greater than this spring force. The static friction force can be calculated using the formula:
[ F_{friction} = \mu_s \times N ]
where ( \mu_s ) is the coefficient of static friction and ( N ) is the normal force. For an object lying on a horizontal surface, the normal force is equal to the gravitational force acting on the object, which can be calculated as:
[ N = m \times g ]
where ( m ) is the mass of the object (12 kg) and ( g ) is the acceleration due to gravity (approximately 9.8 m/s^2). Thus,
[ N = 12 \times 9.8 = 117.6 , \text{N} ]
Since ( F_{friction} = F_{spring} ), we can set the two equations equal to each other and solve for ( \mu_s ):
[ \mu_s \times N = F_{spring} ]
[ \mu_s = \frac{F_{spring}}{N} ]
[ \mu_s = \frac{2}{117.6} ]
[ \mu_s \approx 0.017 ]
Therefore, the minimum value of the surface's coefficient of static friction required to keep the object stationary is approximately 0.017.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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