A toy chest and its contents have a combined weight of W = 180 N. The coefficient of static friction between toy chest and floor μs is 0.440. The child in Fig. 6-35 attempts to move the chest across the floor by pulling on an attached rope. (a) If θ is 42.0°, what is the magnitude of the force that the child must exert on the rope to put the chest on the verge of moving? Determine (b) the value of θ for which F is a minimum and (c) that minimum magnitude.
what is the combined weight of 6.35 and 2.5?
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To solve this problem, we will break it down into three parts.
Part (a): Finding the magnitude of the force the child must exert on the rope to put the chest on the verge of moving.
1. Draw a free-body diagram:
- Draw a vector representing the weight of the toy chest (180 N) pointing downward.
- Draw a vector representing the normal force (N) perpendicular to the floor.
- Draw a vector representing the force of friction (Ff) opposite to the direction of motion.
- Draw a vector representing the force exerted by the child on the rope (F) at an angle of 42.0°.
| /|
| / |
| N / |
|/---/------------->
F Ff W
(42°)
2. Break down the force on the rope:
- Resolve the force exerted by the child on the rope (F) into its components:
Fx = F * cos(θ)
Fy = F * sin(θ)
3. Apply Newton's second law in the vertical direction:
ΣFy = N - W = 0
N = W
The normal force is equal to the weight of the toy chest.
4. Apply Newton's second law in the horizontal direction:
ΣFx = Fx - Ff = 0
Fx = Ff
The force exerted by the child in the horizontal direction is equal to the force of friction.
5. Determine the coefficient of static friction:
μs = Ff / N
Ff = μs * N
Ff = 0.440 * 180 N
6. Solve for the force exerted by the child on the rope:
Fx = Ff
F * cos(θ) = 0.440 * 180 N
F = (0.440 * 180 N) / cos(θ)
Substitute θ = 42.0° and solve for F.
Part (b): Finding the value of θ for which F is a minimum.
1. Find the derivative of F with respect to θ:
dF/dθ = - (0.440 * 180 N) * sin(θ) / (cos^2(θ))
2. Set the derivative equal to zero and solve for θ:
dF/dθ = 0
- (0.440 * 180 N) * sin(θ) / (cos^2(θ)) = 0
Solve for θ.
Part (c): Finding the minimum magnitude of F.
1. Substitute the value of θ obtained in part (b) into the equation for F:
F = (0.440 * 180 N) / cos(θ)
Calculate F for the value of θ obtained in part (b).
Wc = mg = 180N.
Fc = (180N,0Deg.).
Fp = 180sin(0) = o = Force parallel to floor.
Fv = 180cos(0) = 180N. = Force perpendicular to floor.
Ff = u*Fv = 0.44 * 180 = 79.2N. = Force
of static fricion.
a. Fn = Fap*cos42 - Ff = 0,
Fap*cos42 - 79.2 = 0,
Fap*cos42 = 79.2,
Fap = 79.2 / cos42 = 106.6N. = Force
applied.
b. Fap = min. when theta = 0 deg.
c. Fap*cos(0) - 79.2 = 0,
Fap*cos(0) = 79.2,
Fap = 79.2 / cos(0) = 79.2N. = Force
applied. = Min. magnitude.