A worker pushes horizontally on a 35.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the

Question

3 years ago

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A worker pushes horizontally on a 35.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the

crate and the floor is 0.370. (a) What is the value of fs,max under the circumstances? (b) Does the crate move? ("yes" or "no") (c) What is the frictional force fr on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull fpv that will allow the first worker's 112 N push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull fpg that will get the crate moving?

Physics

1 answer:

OLga [1] · Answer 1 · 2022-03-12T00:27:28+03:00

Answer: a) -127 N b) No. c) -112 N d) 40 N e) 15 N

Explanation:

a) Friction force always oppose to the relative movement between two surfaces, and, provided that be less than the fs max, adopt any value to counteract the applied force.

The fs max, is the horizontal component of the contact force, and can be written as follows:

Fs max = us . Fn

As the block is at rest in the vertical direction, this means that Fn must be numerically equal to the weight of the object:

Fn = m g = 35 kg. 9.8 m/s2 = 343 N → Fs max = 0.37. 343 N = 127 N

b) Now, as the applied force is smaller than Fs max, this means that the friction force, is equal and opposite to the applied forcé, i.e., -112 N, so the crate doesn´t move.

c) Please see above.

d) As explained above, the maximum friction force, is proportional to the normal force, which adopts any value needed to satisfy the Newton´s 2nd Law.

So, if we diminish the normal force, we can lower the máximum friction force, helping to the worker to move the crate.

The mínimum needed normal force, will be the one that satisfies the following:

Fs max = F applied = us Fn = 112 N = 0.37. Fn.

Solving for Fn, we get Fn= 303 N

So, the difference between the original normal force and the new one, will be the mínimum upward force needed to make the crate to move, as follows:

Fup = 343 N -303 N = 40 N

e) If we keep the normal force unchanged, but add an horizontal force to help the worker, we will need that the sum of both forces, will be equal to the Fs max, as follows:

Fh + Fapp = 127 N → Fh = 127 N – 112 N = 15 N