A toy chest and its contents have a combined weight of W = 210 N. The coefficien

Question

A toy chest and its contents have a combined weight of W = 210 N. The coefficient of static friction between toy chest and floor mu_s is 0.450. The child in the figure attempts to move the chest across the floor by pulling on an attached rope, (a) If theta is 45.0 degree, 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? Determine (b) the value of theta for which F is a minimum and (c) that minimum magnitude. (a) Number units (b) Number units (c) Number units

Explanation / Answer

(A) in vertical,

Fnet = N + Fsin45 - W = 0

N = W - F sin45

friction force = us N = 0.450 W - 0.450 F sin45

in horizontal,

Fnet = F cos45 - f = 0

F (cos45 + 0.450 sin45) - 0.450 W = 0

F = (0.450 x 210) / (cos45 + 0.450 sin45)

F = 92.2 N ..........Ans

(B) F = 94.5 / (cos(theta) + 0.450 sin(theta))

dF/ d(theta) = - 94.5( - sin(theta) + 0.450 cos(theta)) / ((cos(theta) + 0.450 sin(theta)) )^2

F will be minimum at theta for dF/d(theta) = 0

sin(theta) = 0.450 cos(theta)

theta = tan^-1(0.450)

theta = 24.2 deg   

(c) F = 94.5 ( cos24.2 + 0.450 sin24.2)

F= 86.2 N

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