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 Sep 26th 2019, 07:04 PM #1 Junior Member   Join Date: Sep 2019 Posts: 2 Dynamics Question Hi everyone. I really need help with this problem. I'm not sure if I'm getting the right answer (don't have the answer). Just need to know if my setup is right. Also, do I need to just solve for (ax)? Would that be the only answer I need? Here is the question: An object of mass 15 kg moves on a horizontal surface under the action of a constant force of 100 Newtons at 25 degrees with the surface. The coefficient of kinetic friction between the surface and the object is 0.12. Calculate the acceleration of the object. I attached my work. Thank you for your help!! Attached Thumbnails
 Sep 26th 2019, 07:32 PM #2 Senior Member   Join Date: Apr 2017 Posts: 513 Two possible answers to this .... "a constant force of 100 Newtons at 25 degrees with the surface...." They don't say whether the force is pushing upwards at 25 Deg or downwards at 25 deg to the horizontal .... ======= Please, this is a basic force problem, not an alien one! -Dan neila9876 likes this. Last edited by topsquark; Sep 26th 2019 at 08:56 PM. Reason: Pruning in the name of standard Physics
 Sep 27th 2019, 07:00 AM #3 Junior Member   Join Date: Sep 2019 Posts: 2 So is there no way to solve this? Can anyone tell me if my setup is at least right? I seem to be getting different answers every time I calculate the problem. I'm not sure whether to use -9.80 or +9.80 for gravity.
 Sep 27th 2019, 07:45 AM #4 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,060 As Oz indicates it is unclear if the force is acting downward such that the vertical portion is pushing down in addition with gravity (which will increase the friction) or upward such that the vertical component is lifting the block against gravity (which will decrease friction) Your basic method seems to be OK, but you are perhaps trying to chuck everything into the initial equation and as a result you are loosing track of which way various terms are acting. Try preparing some of the inputs, to the basic A=F/M, separately first and then feed them to the final equation. Thus: HF=Horizontal component of the 100N Force VF=Vertical component of the 100N Force GF=Force due to Gravity FF=Friction Force (which is a function of VF and GF) Is the Friction Force a function of the speed? Take care to ensure that each of these forces is appropriately aligned with the other forces (the sign is acting the right way). It does not matter particularly which way you decide is + and which way is - as long as you keep the same decision at all times. Once you have all the Forces aligned and combined, the rest is trivial. topsquark and benit13 like this. __________________ ~\o/~ Last edited by Woody; Sep 27th 2019 at 07:48 AM.
 Sep 27th 2019, 08:31 AM #5 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 474 Judging from your answer... it seems that the weight and the y-component of the applied force are in the same direction; vertically downwards. I think your answer is wrong because you've put some negative signs on your components that shouldn't be there. Here's a full solution to explain. Assuming that the applied force is pointing downwards by 25 degrees from the horizontal, instead of upwards, we have the following forces labelled on the attached free-body diagram: 1. Weight = $\displaystyle mg$ 2. Reaction force, $\displaystyle R$ 3. Friction force, $\displaystyle F_r \le \mu R$, 4. Applied force = $\displaystyle F_{app}$ = 100 N Let's consider the y-direction. Since we know that the object accelerates along the surface, we know that the reaction force must balance the weight and the applied force. Therefore, according to Newton's first law, the reaction force balances the sum of the weight and y-component of the applied force: $\displaystyle R = mg + F_{app} sin(25)$ Let's now consider the x-direction. In this direction we know the object accelerates along the surface. Therefore, we can solve for Newton's second law along the x-direction. The total force in the x-direction is the difference between the x-component of the applied force and the friction force, which always resists motion: $\displaystyle F_{total,x} = ma = F_{app} cos(25) - F_r$ Because the object is moving, we know the limiting friction force has been overcome, so we know that the friction force relates to the reaction force exactly using: $\displaystyle F_r = \mu R$ We can now substitute these quantities in the second equation and solve for acceleration: $\displaystyle ma = F_{app} cos(25) - \mu R$ $\displaystyle = F_{app} cos(25) - \mu (mg + F_{app} sin(25))$ $\displaystyle a = \frac{F_{app} cos(25) - \mu (mg + F_{app} sin(25))}{m}$ Substituting values: $\displaystyle a = \frac{100 \times cos(25) - 0.12 \times (15 \times 9.81 + 100 \times sin(25))}{15} = 4.527 m/s^2$ Attached Thumbnails   topsquark likes this.

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