The force of friction moving up a slope

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A grocery cart is pushed with a force of 450 N at a 30-degree angle, and the mass of the groceries is 42 kg. To calculate the force of friction with a coefficient of 0.60, the total downward force must be considered, which includes both the gravitational force and the vertical component of the applied force. The downward force is calculated as 636.6 N, combining the weight of the groceries and the vertical component of the pushing force. The correct force of friction is derived from this total downward force, leading to the answer of approximately 380 N. A diagram helped clarify the components of the forces involved in the problem.
tyler24_11
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A grocery cart is being pushed with a force of 450 N at an angle of 30.0 degrees to the horizontal. If the mass of the groceries is 42 Kg,
(a) Calaculate the force of friction if the coefficient is 0.60.

Alright guys I have been doing this problem for around two hours. My physics teacher isn't a good teacher and I have no idea how to get the right answer(located in the back of book).Ive tried all I can think of,
Ff= (0.60)(450n)
= 270n

450 COS 30
= 389.7 N
and a bunch of othe stuff I erased off my paper

The answer in the back is 3.8 x 102 or around 380 n
 
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What is the mass of the cart and groceries?
 
it doesn't say just assume 42kg I geuss
 
tyler24_11 said:
it doesn't say just assume 42kg I geuss

The 450 N force has 2 components. There is the Cos30 component forward, but there is also the Sin30 component down. Sin30*450 N + 42*(9.8) N = 225 + 411.6 N = 636.6 N in the downward direction.

Then you use your coefficient of friction based on the total downward force.
 
yah, thanks. I figured it out last night. I drew a diagram
 

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