55306
A dockworker is loading 20.2-kg crates onto a ship. He notices that it takes 77.2 Newtons of horizontal force to set them into motion from rest. Once in motion, it takes 58.3 Newtons of horizontal force to keep them moving at a constant speed. Determine the coefficient of static friction. Enter your answer accurate to the third decimal place.
Answer =
As the force applied to move the body over the surface, the resisting frictional force is developed between the surfaces. This frictional force is equal to the force applied if the force applied is small. As the applied force is increased by small amounts the frictional force also increases accordingly till the limiting frictional force. If the force is slightly greater then the limiting friction, the body starts sliding. As soon as the body starts sliding the friction decrease and remains constant as the body moves. This is called kinetic friction. The ratio of limiting friction and normal reaction is called coefficient of static friction and that for kinetic friction and normal reaction is coefficient of kinetic friction. If the body is mowing with constant velocity then kinetic friction is equal to the force applied.
If the mass of a body is m then the weight is mg and the normal reaction due to a horizontal surface is mg. Hence
ms = Flim /mg and mk = Ff/mg
Now in our problem ms = Flim /mg = 77.2/(20.2*9.8) = 0.38997 and
mk = Ff/mg = 58.3/(20.2*9.8) = 0.2945.
(Referring to the previous problem.) Determine the coefficient of kinetic friction. Enter your answer accurate to the third decimal place.
Answer =
The coefficient of kinetic friction mk = Ff/mg = 58.3/(20.2*9.8) = 0.2945.
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A box rests on the back bed of a truck. The coefficient of friction between the box and the surface is .789. Calculate the maximum acceleration (in m/s/s) that the truck can have before the box will slide along the truck bed.
Answer = m/s/s
The friction force always tries to avoid relative motion between the surfaces and hence in this case it is in forward direction on the box and backward direction on the truck bed. This forward friction on the box accelerates it. If the box is just not slipping, acceleration ob both will be the same and the friction is limiting friction. Hence
ms = Flim /mg = mamax/mg =amax/g gives
amax= ms g = 0.789*9.8 =7.7322 m/s/s
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Consider the free-body diagram ?? shown at the right. If the applied force is 58.1 N at an angle of 23 degrees, the force of gravity is 127.2 N and the coefficient of friction is .626, then what is the acceleration (in m/s/s) of the object. Is it a negative value for a leftward acceleration?
Answer = m/s/s
Where is your diagram? Still I think it is similar to the problem below and you will be able to solve it. I am not getting it completely.
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A box of books weighing 304 N is shoved across the floor of an apartment by exerting a force of 443 N downward at angle of 35.2 degrees below the horizontal . If the coefficient of kinetic friction is .572, how long (in seconds) does it take the box to move 2.99 meters.
Answer = seconds
If the force applied is not horizontal but making an angle above or below the vertical, we can resolve the force in horizontal and vertical direction. The horizontal component will pull the body and the vertical component will be added to normal reaction. With this change in normal reaction the limiting friction force will also change. Balancing vertical forces we have
N = mg + F sinq =304 + 443 sin35.20 = 559.36N
The acceleration of the box a is given by
F cos q – m N = ma.
Or 443 cos35.2 – 0.572*559.36 = (304/9.8)*a
Or 361.995 – 319.95 = 31.02*a gives
a = 1.355 m/s/s
Now with this acceleration the time taken to cover the distance is given by
X = ut + ½ at2.
Or 2.99 = 0 + 0.5*1.355*t2. gives
t = 2.10 seconds.
A man doing his spring cleaning pulls a 151-N vacuum cleaner across the floor at a constant velocity of 1.13 m/s by exerting a force on it at an angle of 35.4 degrees above the horizontal. If he must pull with a force of 35.5 N to move the vacuum cleaner, what is the coefficient of friction between the vacuum cleaner and the floor? Enter your answer accurate to the fourth decimal place.
Answer =
The problem is similar to the above, the only difference is that the vertical component of the force applied is upward. So
N + F sinq = mg or N = mg - F sinq = 130.4355 N
And F cos q – m N = m0. {constant velocity}
Gives m = F cosq/N = 0.2218.
A student slides a 1323 N crate across the floor by pulling with a force of 761 N at an angle of 30.8 degrees with the horizontal. If the crate moves at a constant speed, find the coefficient of sliding friction between the crate and the floor. Enter your answer accurate to the fourth decimal place.
Answer =
The problem is similar to the above; so
N + F sinq = mg or N = mg - F sinq =933.3354 N
And F cos q – m N = m0. {Constant velocity}
Gives m = F cosq/N = 0.70035.