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Introductory Physics Homework Help
What Determines the Acceleration of a Box on a Moving Plank?
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[QUOTE="asaspades, post: 4527779, member: 490179"] Here are all the problems. I'll post them with my new attempts and then maybe someone can tell me if and where I am going wrong. [B](a) Assume that the coefficient of static friction between the board and the box is not known at this point. What is the magnitude, and the direction, of the acceleration of the box in terms of [itex]F_{\text{f}}[/itex]?[/B] If [itex]a_{\text{box}}[/itex] is the acceleration on the box, then [itex]F_{\text{f}}=m_{1}a_{\text{box}}[/itex] so [itex]a_{\text{box}}=\frac{F_{\text{f}}}{m_{1}}[/itex] This must act in the direction of [itex]F[/itex] or the box would slide or accelerate off the negative side of the board (taking the direction of [itex]F[/itex] to be positive) [B](b) Now take the coefficient of static friction to be [itex]\mu_{S}[/itex]. What is the largest possible magnitude of acceleration of the box? Express your answer in terms of some or all of the variables [itex]F, \mu_{S}, m_{1}, m_{2}[/itex] and [itex] g[/itex].[/B] So [itex] m_{1} a_{\text{box}} = F_{\text{f}} \leq \mu_{S} F_{n}[/itex] where [itex] F_{n} [/itex] is the normal force on the box. [itex] F_{n} = m_{1}g [/itex] so [itex] m_{1} a_{\text{box}} \leq \mu_{S} m_{1}g[/itex] [itex]a_{\text{box}} \leq \mu_{S}g[/itex] and therefore [itex]\max{a_{\text{box}}} = \mu_{S}g[/itex] [B](c) Write down the sum of all horizontal forces on the board, taking the positive direction to be towards the right. Give your answer in terms of [itex]F[/itex] and [itex]F_{f}[/itex].[/B] The total force on the board, [itex]F_{\text{board}}[/itex], is equal to the accelerating force, [itex]F[/itex], plus the friction from the box acting on the board, [itex]-F_{f}[/itex] i.e. [itex]F_{\text{board}}=F-F_{f}[/itex] [B](d) Find an expression for the acceleration of the board when the force of static friction reaches its maximum possible value in terms of [itex]F, \mu_{S}, m_{1}, m_{2}[/itex] and [itex] g[/itex].[/B] [itex]\max{F_{\text{f}}}=\mu_{S}m_{1}g[/itex] so the total force on the board when static friction is at its maximum is [itex]F_{\text{board}}=F-\mu_{S}m_{1}g[/itex]. Because the mass of the box [itex]m_{1}[/itex] acts through the board, [itex]F_{\text{board}}=(m_{1}+m_{2})a_{\text{board}}[/itex], where [itex]a_{\text{board}}[/itex] is the acceleration of the board. This means [itex](m_{1}+m_{2})a_{\text{board}} = F-\mu_{S}m_{1}g[/itex] so [itex]a_{\text{board}} = \frac{F-\mu_{S}m_{1}g}{m_{1}+m_{2}}[/itex] [B](e) What is the minimum value of the constant force, [itex]F[/itex], applied to the board, needed to ensure that the accelerations satisfy [itex]|a_{\text{board}}|>|a_{\text{box}}|[/itex]? Express your answer in terms of some or all of the variables [itex]\mu_{S}, m_{1}, m_{2}, g[/itex] and [itex]L[/itex]. Do [I]not[/I] include [itex]F_{\text{f}}[/itex] in your answer.[/B] Assuming that the force of static friction is at its maximum when slippage occurs, [itex]\left|\frac{F-\mu_{S}m_{1}g}{m_{1}+m_{2}}\right|> |\mu_{S}g|[/itex] If [itex]F<\mu_{S}m_{1}g[/itex] then [itex]\mu_{S}m_{1}g - F > \mu_{S}m_{1}g + \mu_{S}m_{2}g [/itex] then [itex]-F > \mu_{S}m_{2}g[/itex] and [itex] F < -\mu_{S}m_{2}g[/itex]. This implies F is negative when it is actually positive so [itex]F \geq \mu_{S}m_{1}g[/itex] and then [itex]F-\mu_{S}m_{1}g > \mu_{S}m_{1}g + \mu_{S}m_{2}g [/itex] and [itex]F>\mu_{S}g(2m_{1}+m_{2})[/itex] [/QUOTE]
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What Determines the Acceleration of a Box on a Moving Plank?
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