a) A crane of mass M has a wheelbase, ?, and its centre of mass midway between t

Question

a)

A crane of mass M has a wheelbase, ?, and its centre of mass midway between the wheels (the mass of the boom is negligible). The boom has length L and makes an angle ? with the horizontal. The crane is in contact with the ground only at its front and rear tires. The diagram shows the crane, supporting a mass m in front.

We will be considering two aspects: (a) the force of the rear wheels on the ground, and (b) when the crane tips over, as a function of m and ?. Answer True or False to each of the statements below; e.g., if the first statement is true and the rest, false, enter TFFFF. You only have 5 tries!

1)The crane will always tip if m > M, for any choice of ?.

2)The crane is in static equilibrium. It is most convenient to calculate torques about the point where the front wheel touches the ground.

3)As shown, the weight of the crane leads to a clockwise torque about the rear axle.

4)As shown, the force of the front wheel on the ground is equal to Mg/2.

5)If the crane tips due to the mass m, the front wheel would remain in contact with the ground.

If the crane does not tip, which of the equations gives the force of the rear wheels on the ground? E.g., enter A. You only have 3 tries!

b)

If the crane does not tip, which of the equations gives the force of the rear wheels on the ground?

N = g(M/2?(mLsin?)/?)

N = g(m(1/2?(mLcos?)/?))

N = g((M+m)/2?(mLcos?)/?)

N = g(M/2?(mLcos?)/?)

N = g(m(1/2?(mLsin?)/?))

N = g((M+2m)/2?(mLcos?)/?)

N = g((M+m)/2?(m?cos?)/L)

N = g((M+2m)/2?(mLsin?)/?)

None of these

c)

The angle is adjusted so that the crane could tip. What is the maximum value of m before the crane will tip at that angle? You only have 3 tries!

None of these

d)What is the maximum load the crane can lift if ?=0°? Assume M = 2.10×104 kg, ? = 5.50 m, and L = 17 m.

A) m = M?/(2Lcos??2?) B) m = M?/(2Lsin??2?) C) m = M?/(2Lcos???) D) m = ML/(2?cos?) E) m = M?/(2Lsin???) F) m = M?/(2Lsin?) G) m = M?/(2Lcos?) H)

None of these

Explanation / Answer

1)

FALSE

2)

FALSE

3)

TRUE

4)

FALSE

5)

TRUE

b)

net rorque about the fron wheel is zero

M*g*l/2 = N*l + m*g*L*costheta

N = M*g*1/2 - m*g*L*costheta/l

N = g*( M/2 - m*L*costheta/l )

if N = 0

M*g*1/2 - m*g*L*costheta/l = 0

M*g*1/2 = m*g*L*costheta/l

M/2 = m*L*costheta/l

m = M*l/(2L*costheta)

option (D)

d)

m = 2.1*10^4*5.5/(2*17*cos0)

m = 3397 kg <<------answer

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