A toy chest and its contents have a combined weight of 180 N. The coefficient of static friction between toy chest and floor is 0.42.The child in Fig. 6-38 attempts to move the chest across the floor by pulling on an attached rope. (a) If θ is 42°, what is the magnitude of the force F that the child must exert on the rope to put the chest on the verge of moving? (b) Write an expression for the magnitude F required to put the chest on the verge of moving as a function of the angle θ. Determine (c) the value of θ for which F is a minimum and (d) that minimum magnitude.
The Correct Answer and Explanation is :
Given:
- The combined weight of the toy chest and its contents = 180 N.
- The coefficient of static friction, ( \mu_s = 0.42 ).
- The angle at which the child pulls the rope, ( \theta = 42^\circ ).
(a) Magnitude of the force ( F ) to put the chest on the verge of moving:
The force ( F ) that the child exerts can be broken into two components:
- The horizontal component: ( F_{\text{horizontal}} = F \cos \theta ).
- The vertical component: ( F_{\text{vertical}} = F \sin \theta ).
The weight of the chest, ( W = 180 \, \text{N} ), creates a normal force. The normal force, ( N ), is affected by both the weight of the chest and the vertical component of the applied force:
[
N = W – F \sin \theta.
]
The force of static friction is given by:
[
f_{\text{friction}} = \mu_s N = \mu_s (W – F \sin \theta).
]
For the chest to be on the verge of moving, the applied horizontal force must balance the static friction:
[
F \cos \theta = \mu_s (W – F \sin \theta).
]
Solve for ( F ):
[
F \cos \theta + F \mu_s \sin \theta = \mu_s W.
]
[
F (\cos \theta + \mu_s \sin \theta) = \mu_s W.
]
[
F = \frac{\mu_s W}{\cos \theta + \mu_s \sin \theta}.
]
Substituting ( \mu_s = 0.42 ), ( W = 180 \, \text{N} ), and ( \theta = 42^\circ ), we can calculate the magnitude of ( F ).
(b) Expression for ( F ) as a function of ( \theta ):
From the above, the general expression for the magnitude of the force is:
[
F(\theta) = \frac{\mu_s W}{\cos \theta + \mu_s \sin \theta}.
]
(c) Angle ( \theta ) for which ( F ) is minimum:
To minimize ( F ), we differentiate ( F(\theta) ) with respect to ( \theta ) and set the derivative equal to zero:
[
\frac{dF}{d\theta} = 0.
]
This gives us the value of ( \theta ) where ( F ) is minimized.
(d) Minimum magnitude of ( F ):
Substitute the value of ( \theta ) from part (c) into the expression for ( F(\theta) ) to find the minimum force required.
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