Physics Homework Help: Calculating Frictional Force and Tension on a Slope

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To calculate the frictional force and normal reaction on a slope, first determine the normal force, which acts perpendicular to the slope. For a 1.5 kg object on a 30-degree slope, the normal force can be calculated using the weight component acting perpendicular to the slope. Since the object is at rest, the net force is zero, indicating that static friction balances the other forces. The frictional force can be found by multiplying the coefficient of friction by the normal force. For the tugboat problem, tension can be calculated by balancing the forces acting on the ship against the drag force.
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frictional force! help!

Hi There,
I'm having trouble with these physics question in my homework:

1. An object of mass 1.5kg rests on a slope making an angle of 30 degrees, with the horizantal calculate:

a) the frictional force and
b) the normal reaction acting on the object. Gravitational field strengh = 9.8N/kg

2. Two tug boats pull a ship so that it moves with a constant velocity. Each of the tugs produces a tension T at an angle of 28 degrees to the direction the ship is moving. The drag on the ship is 220kN. Calculate T.

I would be really grateful if you go could help me with this, its driving me nuts...oh and by the way with question 1, I worked out the normal force, which I think is what I had to do :confused: and got 1.3N (1dp)...after that I got stuck..especially with the whole coefficent friction
 
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Okay we'll start with question 1.
What do we know about the normal reaction force?
Well, it's a normal force, which means that it acts at a right angle to the plane it's resting on (the force pushing on the mass).

You know that gravity acts straight down. So you just need to calculate the component of weight (mg) that is perpendicular to the plane.

That sums up b)

For a):

The question states that the object is at rest. This implies that the net force must be zero (and so is initial velocity). Static friction (and kinetic friction) is measured by multiplying a coeffiecient of the friction by the normal reaction force.
Sum up all of the forces acting on the box and let them equal zero (sum up each dimension individually), the only variable you do not have is friction.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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