1. A crane of mass M has a wheelbase, , and its centre of mass midway between th

Question

1. 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!

2. If the crane does not tip, which of the equations gives the force of the rear wheels on the ground? E.g., enter A.

Explanation / Answer

forces acting on the system:

weight of the trolley=M*g, vertically downward

weight of mass m=m*g, vertically downward

normal force from rear wheel=F1, vertically upward

normal force from front wheel=F2, vertically upward

balancing forces:
M*g+m*g=F1+F2

==>F1+F2=(M+m)*g...(1)

answers:

part a: TRUE: weight of the crane has a torque in anticlockwise direction about the front wheel.(toruqe=cross product of distance vector

and force vector. )

part b: FALSE

force on the real wheel will depend upon L, theta , M and m to balance force and torque. hence its value is not fixed at M*g/2.

part c: FLASE

for tipping, torque needs to be unbalanced.

let us consider torque about front wheel:

total torque=m*g*L*cos(theta)-M*g*0.5*l+F1*l

with F1+F2=M*g+m*g

using F1=(M+m)*g-F2

total torque=m*g*L*cos(theta)-0.5*M*g*l+M*g*l+m*g*l-F2*l

=m*g*(L*cos(theta)+l)-(0.5*M*g*l+F2*l)

so if we can make the value of m and L such that this net torque is greater than zero, then the crane will tip

part d:TRUE

part e: TRUE

hence final solution:
TFFTT

part 2:
if static equilibrium exists,

the torque will also be balanced.

hence net torque on front wheel is zero.

m*g*L*cos(theta)-M*g*0.5*l+F1*l=0

==>F1=(0.5*M*g*l-m*g*L*cos(theta)/l

hence option E is correct.

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