Physics Qbank 1: motion, force, and work #17&20
Physics Qbank 1: motion, force, and work #17&20
So I was working on number #17 and 20 which both relate to static friction. On #17 the explanation used mgcos but for #19 they used mgsin? How do we know which direction static friction is going in?

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 Joined: Fri May 25, 2018 9:15 am
Re: Physics Qbank 1: motion, force, and work #17&20
Thanks for the question!
We'll use different trig values for different situations depending on what we're being asked and what information we're given. I've attached the image from the explanation of Q17 which shows how to set up the ramp with trig values, in general: For Q17 question, we basically have to figure out what upward force we have to apply to overcome static friction. Static friction is determined by the coefficient of static friction (0.5) and the normal force (which will point perpendicular to the ramp, and will be equal to the vertical component of gravity, but in the opposite direction). The force down the block, which the force of static friction opposes, will be equal to the horizontal component of gravity, but in the opposite direction.
From our triangle, the horizontal component of gravity (down the plane) is mgsin30. The vertical component of gravity (perpendicular to the plane) is mgcos30. These values are always set this way, which you can prove via similar triangles trigonometry, if you want, but it's easier to memorize!
So gravitational force down the plane is mgsin30 = 50 N. Static friction exerts force µFn = µmg cos 30º (since this is the vertical component of force). We need to overcome this static frictional force to start the movement, so:
Fapp = 50N + µmg cos 30º = approx. 95 N.
Q20 has the same general setup, it just depends which component we're looking at. So, if we're looking at normal force (which we use to determine frictional force), we'll use cos, but if we're looking at the gravitational force down the ramp (parallel to the ramp), we use sin.
The above explanation got a little away from the original question, but I hope it's still helpful! The basic idea with these triangles is captured in the attachment. If we want parallel to the plane of the ramp, we'll use sin. If we want perpendicular to the plane of the ramp, we'll use cos. Note that the angle in these cases is the angle that the ramp makes with the ground. Static and kinetic friction will always be parallel to the plane of the ramp, but friction is in part determined by the normal force, which is perpendicular to the plane of the ramp. There may be situations where we need to use one or the other, or even both!
I hope this is helpful! Good luck with your prep!
We'll use different trig values for different situations depending on what we're being asked and what information we're given. I've attached the image from the explanation of Q17 which shows how to set up the ramp with trig values, in general: For Q17 question, we basically have to figure out what upward force we have to apply to overcome static friction. Static friction is determined by the coefficient of static friction (0.5) and the normal force (which will point perpendicular to the ramp, and will be equal to the vertical component of gravity, but in the opposite direction). The force down the block, which the force of static friction opposes, will be equal to the horizontal component of gravity, but in the opposite direction.
From our triangle, the horizontal component of gravity (down the plane) is mgsin30. The vertical component of gravity (perpendicular to the plane) is mgcos30. These values are always set this way, which you can prove via similar triangles trigonometry, if you want, but it's easier to memorize!
So gravitational force down the plane is mgsin30 = 50 N. Static friction exerts force µFn = µmg cos 30º (since this is the vertical component of force). We need to overcome this static frictional force to start the movement, so:
Fapp = 50N + µmg cos 30º = approx. 95 N.
Q20 has the same general setup, it just depends which component we're looking at. So, if we're looking at normal force (which we use to determine frictional force), we'll use cos, but if we're looking at the gravitational force down the ramp (parallel to the ramp), we use sin.
The above explanation got a little away from the original question, but I hope it's still helpful! The basic idea with these triangles is captured in the attachment. If we want parallel to the plane of the ramp, we'll use sin. If we want perpendicular to the plane of the ramp, we'll use cos. Note that the angle in these cases is the angle that the ramp makes with the ground. Static and kinetic friction will always be parallel to the plane of the ramp, but friction is in part determined by the normal force, which is perpendicular to the plane of the ramp. There may be situations where we need to use one or the other, or even both!
I hope this is helpful! Good luck with your prep!
Re: Physics Qbank 1: motion, force, and work #17&20
Thanks so much!
Re: Physics Qbank 1: motion, force, and work #17&20
How are we able to add the forces of two different components (ie x and y components?)
thanks in adv!

 Posts: 537
 Joined: Sat Mar 30, 2019 8:39 pm
Re: Physics Qbank 1: motion, force, and work #17&20
What will is describing are two components of force parallel to the plane of the slope (those 'pulling' the box down the slope). Those components are found by simply decomposing two other vectors (gravity and static friction) into their components parallel to the slope.