Asked by S.Jacob

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.
Answered by Henry
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.

Answered by smart student
this is wrong
Answered by Anonymous
what is the combined weight of 6.35 and 2.5?
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