In the diagram shown below, a board with mass 24.6 kg and length 2.41 m is suspe
ID: 2097287 • Letter: I
Question
In the diagram shown below, a board with mass 24.6 kg and length 2.41 m is suspended by a taut cable and held horizontally against a vertical wall with ? = 14.6o. A box is placed a distance 0.291 m from the side of the wall. If the coefficient of static friction between the board and the wall is 0.505, what is the maximum mass of the box so that the board can remain in equilibrium? Write your answer to three significant figures in kg. Do not include your units in the space below. (Hint: you will have to think about the force that the wall exerts on the board. It has two components, which component is the normal force? which component can be regarded as the static friction force?)
Explanation / Answer
N + Tsin(#) = Mg + mg ----- (1)
where # in angle and m is unkown
N = Tcos(#)
0.291*mg + 2.41/2*Mg = Tsin(#)*2.41
0.291*mg /2.41 + 2.41/2*Mg/2.41 = Tsin(#)
dividing eq 1 by Tsin(#) and solving for m
m = 17.896 Kg