Friction of an object being pulled up/down

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The discussion focuses on calculating the normal force on an 80kg object being pulled with a force of 4000N at angles of 50 degrees and 70 degrees. For the upward pull at 50 degrees, the calculations suggest a negative normal force, indicating the object would lift off the ground, resulting in no normal force. In contrast, for the downward pull at 70 degrees, the normal force is calculated to be 268.14N. Participants emphasize the importance of drawing a free body diagram and applying Newton's second law for inclined plane problems. Ultimately, the scenario of pulling the object at 50 degrees leads to confusion about the normal force due to the significant pulling force exceeding the weight.
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Homework Statement


Find the normal force on an object, 80kg, which is being pulled by a force of 4000N. A) Up at an angle of 50 degrees, B) Down at an angle of 70 degrees

Homework Equations


Fn = Fwcosx
Fp = Fwsinx

The Attempt at a Solution



A)
Fp = Fwsinx
Fp = 4000sin50 = 3064N
(80*9.8) - 3064 = -2280N = 2280N

B)
Fp = 4000sin70 = 3758.8N
(80*9.8) + 3758.8 = 4542.7N

Is this correct?
 
Last edited:
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I'm confused. Is the object being pulled up an inclined plane, or is it on a flat surface and the applied force acts at the given angles?
 
Pulled up on an inclined plane.
 
In that case your "Relevant Equations" are wrong. You need to draw a free body diagram and apply Newton's 2nd law to it. That's how you do every inclined plane problem.
 
Tom Mattson said:
In that case your "Relevant Equations" are wrong. You need to draw a free body diagram and apply Newton's 2nd law to it. That's how you do every inclined plane problem.

Ah...I believe I was doing the other scenario then. In that case, I would simply do Fn = 80 * 9.8 * cos 50 = 503.9 N and Fn = 80 * 9.8 * cos70 = 268.14N, correct?
 
Last edited:
The other scenario doesn't make much sense. If you pull on an 80 kg object with 4000 N at 50 degrees, the object will be lifted right off the ground. Hence, no normal force.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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