A truck carries a heavy crate. The coefficient of static friction between the body and the bed of the truck is 0.73. What is the maximum speed at which the driver can decelerate while preventing the box from sliding against the cab?
Answer 1
Since we don’t want the crate to slip, we need to see when the deceleration force equals the friction force. Leave
m = mass of the box
a = acceleration (in this case deceleration of the truck)
g = acceleration due to gravity, 9.8 m/s^2
μ = coefficient of friction, 0.73
So
ma = µmg
a = µg
a = 0.73 * 9.8 m/s^2
a = 7.154 m/s^2
So the driver can safely decelerate at about 7.1 m/s^2, harder, and the box is likely to slide against the cab.
answer 2
A flatbed truck is loading
answer 3
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To prevent the crate from sliding (using Newton’s 2nd law of motion), F = ma where F = force acting between the crate and the bed Note that the only force acting between the crate and the bed is the force of friction and it is F = – µmg where µ = coefficient of friction m = mass of the box g = acceleration due to gravity = 9.8 m/s^2 NOTE that the negative sign in F is simply an indication that the force is acting in the direction opposite to the movement of the carriage. Equating the two equations for F, – µmg = ma and as “m” appears on both sides of the equation, it cancels and the above simplifies to a = – µg and replacing the values a = – 0 .75(9.8) ) a = – 7.35 m/s^2 Hope this helps.