What is the maximum possible magnitude of the force of static friction between t

Question

What is the maximum possible magnitude of the force of static friction between this table and this box (with this specific table top orientation)?

QUESTION 11 Question Completion Status: Consider the following situation: = 0.3 = 0.2 = 15° kg box on an inclined A 2 table. Please follow the instructions for this homework carefully. This problem and the next three pertain to the same physical situation. List the forces acting on the box (we will consider these forces acting on the box as the "action” forces): Force #1 Force #2 Force #3 The net force acting on the box has magnitude of and is directed The acceleration of the box has magnitude of and is directed To what angle (to the horizontal) would you need to increase the tilt of the table top in order for the box to start sliding? (round to two decimals) The following three problems contain questions about forces you listed above as #1 and #3 with the table inclined at the angle of 15° to the horizontal. degrees.

Explanation / Answer

force1: Weight force (gravitational force by earth)

force2: Normal force by table

force3: friction force by table

perpendicular the incline,

Fnet = N - m g cos15= 0

N = m g cos15

f_max = us N = 0.3 x 2 x 9.8 x cos15 = 5.68 N

component of weight force along the incline,

= 2 x 9.8 x sin15 = 5.07

this is less than maximumvalue of friction.

hence block will not move.

Fnet = 0 ....Ans

a = 0 ......Ans

to start moving,

0.3 x m g cos(theta) = m g sin(theta)

tan(theta) = 0.3

theta = 16.7 deg .....Ans

maximum magnitude of friction = 5.68 N .....Ans

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