A worker applies a horizontal force to the top edge of a crate to get it to tip
ID: 1376716 • Letter: A
Question
A worker applies a horizontal force to the top edge of a crate to get it to tip forward (see the figure (Figure 1) ).
Part A
If the create has a mass of 60kg and is 1.6 m tall and 0.80 m in depth and width, what is the minimum force needed to make the crate start tipping? (Assume the center of mass of the crate is at its center and static friction great enough to prevent slipping.)
Express your answer using two significant figures.
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Figure 1 of 1
A worker applies a horizontal force to the top edge of a crate to get it to tip forward (see the figure (Figure 1) ).
Part A
If the create has a mass of 60kg and is 1.6 m tall and 0.80 m in depth and width, what is the minimum force needed to make the crate start tipping? (Assume the center of mass of the crate is at its center and static friction great enough to prevent slipping.)
Express your answer using two significant figures.
F= ____ NSubmitMy AnswersGive Up
Incorrect; Try Again; 4 attempts remaining
Provide FeedbackContinue
Figure 1 of 1
A worker applies a horizontal force to the top edge of a crate to get it to tip forward (see the figure (Figure 1) ). Part A If the create has a mass of 60kg and is 1.6 m tall and 0.80 m in depth and width, what is the minimum force needed to make the crate start tipping? (Assume the center of mass of the crate is at its center and static friction great enough to prevent slipping.) Express your answer using two significant figures. SubmitMy AnswersGive Up Incorrect; Try Again; 4 attempts remaining Provide FeedbackContinueExplanation / Answer
T1 = torque by the force
T2 = torque by gravity at centre of mass
T1 + T2 = 0
T1 = F*sin(theta)*d
T2 = - mg*cos(theta)*d /2
tan(theta) = 1.6/0.80 = 2
theta = 63.4 deg
0 = T1 +T2
0 =(F*sin(theta)*d) + (- mg*cos(theta)*d/2)
F = ( 60*9.8*cos(63.4)*0.8) / (sin(63.4)*1.6)
F = 147.224 N
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