The cart and its contents have a total mass of 4500 kg and a center of gravity at G as shown in Figure Q4. The coefficient of static friction between the wheels and the tracks is 4, = 0.4 when the wheels are locked. The normal forces acting on the wheels at (A and B) are independent as to whether the wheels are locked or not. Therefore, 10kN G 0.9 m V0.15 m : 0.6 m 1.5 m Figure 04: The cart

International Edition---engineering Mechanics: Statics, 4th Edition
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Author:Andrew Pytel And Jaan Kiusalaas
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Chapter7: Dry Friction
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Problem 7.73P: The normal pressure acting on the disk of the sander is given by p=(4/3)+(r2/6), where p is the...
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The cart and its contents have a total mass of 4500 kg and a center of gravity at G as shown in
Figure Q4. The coefficient of static friction between the wheels and the tracks is µ, = 0.4 when
the wheels are locked. The normal forces acting on the wheels at (A and B) are independent as
to whether the wheels are locked or not. Therefore,
10kN
G
0.9 m
A20.15 m
: 0.6 m
1.5 m
Figure Q4: The cart.
(a)
Draw the free-body diagram of the system.
(b)
Calculate the normal force acting on the front wheels at B when the brakes at both A and
B are locked.
Transcribed Image Text:The cart and its contents have a total mass of 4500 kg and a center of gravity at G as shown in Figure Q4. The coefficient of static friction between the wheels and the tracks is µ, = 0.4 when the wheels are locked. The normal forces acting on the wheels at (A and B) are independent as to whether the wheels are locked or not. Therefore, 10kN G 0.9 m A20.15 m : 0.6 m 1.5 m Figure Q4: The cart. (a) Draw the free-body diagram of the system. (b) Calculate the normal force acting on the front wheels at B when the brakes at both A and B are locked.
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