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user14245
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The problem
A [itex]0.250 \operatorname{kg}[/itex] block of cheese lies on the floor of a [itex]900 \operatorname{kg}[/itex] elevator cab that is being pulled upward by a cable through distance [itex] d_1 = 2.40 \operatorname{m}[/itex] and then through distance [itex]d_2=10.5 \operatorname{m}[/itex]. Through [itex]d_1[/itex], if the normal force on the block from the floor has constant magnitude [itex]F_N = 3.00\operatorname{N}[/itex], how much work is done on the cab by the force from the cable?
From Fundamentals of Physics, 9th Editon, Problem 25, Chapter 7
My solution manual
Here is what my solution manual says:
The net upward force is given by [itex]F + F_N-(m+M)g = (m+M)a[/itex] where [itex]m=0.250 \operatorname{kg}[/itex] is the mass of the cheese, [itex]M = 900 \operatorname{kg}[/itex] is the mass of the elevator cab, [itex]F[/itex] is the force from the cable, and [itex]F_N = 3.00 \operatorname{N}[/itex] is the normal force on the cheese. On the cheese alone, we have [itex]F_N - mg = ma[/itex], and etc..., the solution continues
My question
I do not see why the first equation is correct. To me, the force [itex]F_N[/itex] is internal between the cab and the block, so once one considers both as a system and applies the Newton's second law, the equation should read [itex]F - (m+M)g = (m+M)a[/itex].
Is the solution manual wrong, or am I overlooking something?
Thank you.
A [itex]0.250 \operatorname{kg}[/itex] block of cheese lies on the floor of a [itex]900 \operatorname{kg}[/itex] elevator cab that is being pulled upward by a cable through distance [itex] d_1 = 2.40 \operatorname{m}[/itex] and then through distance [itex]d_2=10.5 \operatorname{m}[/itex]. Through [itex]d_1[/itex], if the normal force on the block from the floor has constant magnitude [itex]F_N = 3.00\operatorname{N}[/itex], how much work is done on the cab by the force from the cable?
From Fundamentals of Physics, 9th Editon, Problem 25, Chapter 7
My solution manual
Here is what my solution manual says:
The net upward force is given by [itex]F + F_N-(m+M)g = (m+M)a[/itex] where [itex]m=0.250 \operatorname{kg}[/itex] is the mass of the cheese, [itex]M = 900 \operatorname{kg}[/itex] is the mass of the elevator cab, [itex]F[/itex] is the force from the cable, and [itex]F_N = 3.00 \operatorname{N}[/itex] is the normal force on the cheese. On the cheese alone, we have [itex]F_N - mg = ma[/itex], and etc..., the solution continues
My question
I do not see why the first equation is correct. To me, the force [itex]F_N[/itex] is internal between the cab and the block, so once one considers both as a system and applies the Newton's second law, the equation should read [itex]F - (m+M)g = (m+M)a[/itex].
Is the solution manual wrong, or am I overlooking something?
Thank you.