2009 BR Physics Ex 2.12b

[popchap]

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Q: A second back packer grabs her 75 N pack and pulls it with a constant speed at an angle of 30 degrees to the horizontal. The force she applies to the pack is 40N. What is the largest magnitude of frictional force that is acting on the pack?

I don't understand the set up for this question. The solution tells me that Net F in x direction = ma = 0. Then to find the fs the set up was (40N) (cos30) - fs = 0.

Can someone please explain to me in details how to set this problem up? Thanks!

 

[Temperature101]

Q: A second back packer grabs her 75 N pack and pulls it with a constant speed at an angle of 30 degrees to the horizontal. The force she applies to the pack is 40N. What is the largest magnitude of frictional force that is acting on the pack?

I don't understand the set up for this question. The solution tells me that Net F in x direction = ma = 0. Then to find the fs the set up was (40N) (cos30) - fs = 0.

Can someone please explain to me in details how to set this problem up? Thanks!

Do the free body diagram... The friction force (fs) is opposite direction to the force (Fx) pulling the object at that 30 degree ....The total net force in the x-drection.... (Fx - fs = ma) or (40N)cos (30) - fs = ma ....a = 0 since the pulling the object at a constant speed...Fx( X-direction).....cos 30 = adjacent/hypothenuse = (Fx/F)....Fx = F cos (30) = (40N )(cos 30)... That is how they got the force acting in the opposite direction to the frictional force.

 

[popchap]

Thanks so much! It makes sense now.