You are lowering two boxes, one on top of the other, down the ramp shown in the

Q: You are lowering two boxes, one on top of the other, down the ramp shown in the

Here, a) fk + T - mtot*g*sintheta = mtot *a = 0 fk = uk * mtot*g*costheta T = mtot *g(sintheta - uk*costheta) T = ( 2.50/(2.50^2 + 4.75^2)^1/2 - 0.435 * 4.75/(2.50^2 + 4.75^2)^1/2) * (32+48)*9.81 = 63.4173 N b) fs = m*g*sintheta = 32 * 9.81 * 2.50/(2.50^2 + 4.75^2)^1/2 = 146 N

ID: 1424124 • Letter: Y

Question

You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 19.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.435, and the coefficient of static friction between the two boxes is 0.816.

What force do you need to exert to accomplish this?

265 N is not the answer; i have already tried

What is the magnitude of the friction force on the upper box?

226.4 is not the answer for this.

Explanation / Answer

Here,

fk + T - mtot*g*sintheta = mtot *a = 0

fk = uk * mtot*g*costheta

T = mtot *g(sintheta - uk*costheta)

T = ( 2.50/(2.50^2 + 4.75^2)^1/2 - 0.435 * 4.75/(2.50^2 + 4.75^2)^1/2) * (32+48)*9.81 = 63.4173 N

fs = m*g*sintheta = 32 * 9.81 * 2.50/(2.50^2 + 4.75^2)^1/2 = 146 N

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